Chapter 12 ยท Limits and Derivatives

12.2 Limits of Polynomial, Rational and Trigonometric Functions

Polynomials and rational functions: substitute โ€” and when you get $\tfrac00$, factorise or rationalise. Then the result $\displaystyle\lim_{x\to a}\tfrac{x^n-a^n}{x-a}=na^{n-1}$ and, through the sandwich theorem, the two trigonometric limits every derivative of sine and cosine depends on.

In Section 12.1 we guessed limits from tables and graphs. Tables are slow and can mislead, so now we want quick, reliable methods. For most functions you meet in Class 11 the method is short: if the expression makes sense at $x=a$, the limit is simply its value there. The interesting cases โ€” the ones derivatives are made of โ€” are those where substituting $x=a$ gives the meaningless $\tfrac00$. The average speed $\tfrac{s(2+h)-s(2)}{h}$ at $h=0$ is exactly such a case.

In this section we handle $\tfrac00$ for polynomials and rational functions by factorising (cancelling the common factor $x-a$) or by rationalising square roots. We prove the standard result $\displaystyle\lim_{x\to a}\tfrac{x^n-a^n}{x-a}=na^{n-1}$, which will give the derivative of $x^n$. Finally we meet trigonometric limits: with a picture of the unit circle and the sandwich theorem we prove that $\tfrac{\sin x}{x}\to1$ and $\tfrac{1-\cos x}{x}\to0$ as $x\to0$, and use them to evaluate a whole family of limits.

Limits of Polynomials and Rational Functions

A polynomial is built from the constant functions and $x$ by multiplication and addition, so the algebra of limits (Theorem 12.2) applies to every piece of it.

๐Ÿ“ Theorem 12.3 โ€” Limits by substitution

(a) If $p(x)=a_0+a_1x+a_2x^2+\dots+a_nx^n$ is a polynomial, then for every real $a$ $$\lim_{x\to a}p(x)=p(a).$$

(b) If $p$ and $q$ are polynomials and $q(a)\ne0$, then $$\lim_{x\to a}\dfrac{p(x)}{q(x)}=\dfrac{p(a)}{q(a)}.$$

Proof. (a) By Section 12.1, $\displaystyle\lim_{x\to a}x^k=a^k$ for each $k$. The scalar-multiple rule gives $\displaystyle\lim_{x\to a}a_kx^k=a_ka^k$, and the sum rule, used $n$ times, gives $\displaystyle\lim_{x\to a}p(x)=a_0+a_1a+\dots+a_na^n=p(a)$. (b) By (a) the numerator tends to $p(a)$ and the denominator to $q(a)\ne0$, so the quotient rule applies. $\blacksquare$

When $q(a)=0$ the theorem says nothing, and there are two very different situations.

๐Ÿงญ Problem-solving strategy โ€” Evaluating $\displaystyle\lim_{x\to a}\tfrac{p(x)}{q(x)}$
  1. Substitute $x=a$. If the denominator is not $0$, the value is the limit. Stop.
  2. If you get $\tfrac{c}{0}$ with $c\ne0$: there is no finite limit (look at the one-sided behaviour if the question asks).
  3. If you get $\tfrac00$: factorise numerator and denominator, cancel $(x-a)$, and substitute again. Repeat if $\tfrac00$ appears once more.
  4. If square roots are involved, rationalise: multiply top and bottom by the conjugate, e.g. $\sqrt{x}+\sqrt a$, to bring out the factor $x-a$.
  5. If $x^n-a^n$ appears, use Theorem 12.4 below instead of long division.
โœ๏ธ Example 1 โ€” Direct substitution
Evaluate (a) $\displaystyle\lim_{x\to2}(x^3-3x+5)$ (b) $\displaystyle\lim_{x\to-1}\dfrac{2x^2+3}{x-4}$ (c) $\displaystyle\lim_{x\to1}(x^{10}+x^5+1)$ (d) $\displaystyle\lim_{x\to3}\dfrac{x+1}{x-3}$, if it exists.
(a) A polynomial: $8-6+5=7$.
(b) The denominator at $-1$ is $-5\ne0$: $\dfrac{2+3}{-5}=-1$.
(c) $1+1+1=3$.
(d) Substituting gives $\tfrac40$: non-zero over zero, so there is no finite limit. (Just right of $3$ the fraction is large and positive, just left of $3$ large and negative.)
โœ๏ธ Example 2 โ€” Cancelling the common factor
Evaluate (a) $\displaystyle\lim_{x\to3}\dfrac{x^2-5x+6}{x^2-9}$ (b) $\displaystyle\lim_{x\to1}\dfrac{x^3-1}{x^2+2x-3}$ (c) $\displaystyle\lim_{x\to-2}\dfrac{x^3+8}{x+2}$.
(a) At $x=3$: $\tfrac{9-15+6}{9-9}=\tfrac00$. Factorise: $\dfrac{(x-2)(x-3)}{(x-3)(x+3)}=\dfrac{x-2}{x+3}$ for $x\ne3$, so the limit is $\tfrac{1}{6}$.
(b) $\tfrac00$ at $x=1$. $x^3-1=(x-1)(x^2+x+1)$ and $x^2+2x-3=(x-1)(x+3)$, so the limit is $\dfrac{1+1+1}{1+3}=\dfrac34$.
(c) $\tfrac00$ at $x=-2$. Sum of cubes: $x^3+8=(x+2)(x^2-2x+4)$, so the limit is $4+4+4=12$.
Check (b) at $x=1.001$: $\tfrac{0.003003001}{0.004001}\approx0.75056$ โœ“.
โœ๏ธ Example 3 โ€” Rationalising
Evaluate (a) $\displaystyle\lim_{x\to0}\dfrac{\sqrt{1+x}-1}{x}$ (b) $\displaystyle\lim_{x\to4}\dfrac{x-4}{\sqrt x-2}$ (c) $\displaystyle\lim_{x\to0}\dfrac{\sqrt{4+x}-\sqrt{4-x}}{x}$.
(a) Multiply top and bottom by $\sqrt{1+x}+1$: $\dfrac{(1+x)-1}{x(\sqrt{1+x}+1)}=\dfrac{1}{\sqrt{1+x}+1}$ for $x\ne0$. Substitute: $\tfrac12$.
(b) Write $x-4=(\sqrt x-2)(\sqrt x+2)$ (for $x\ge0$). The fraction is $\sqrt x+2$ for $x\ne4$, so the limit is $4$.
(c) Multiply by $\sqrt{4+x}+\sqrt{4-x}$: the numerator becomes $(4+x)-(4-x)=2x$, so the fraction is $\dfrac{2}{\sqrt{4+x}+\sqrt{4-x}}\to\dfrac{2}{2+2}=\dfrac12$.
Check (a) at $x=0.001$: $\tfrac{1.0004999-1}{0.001}\approx0.49988$ โœ“.

The Standard Limit $\dfrac{x^n-a^n}{x-a}$

Expressions like $\tfrac{x^5-32}{x-2}$ are of the form $\tfrac00$ at $x=2$ and can be factorised โ€” but dividing out $x-2$ by hand is tedious. One identity does them all at once.

๐Ÿ“ Theorem 12.4 โ€” The limit of $\dfrac{x^n-a^n}{x-a}$

For every positive integer $n$ and every real number $a$, $$\lim_{x\to a}\dfrac{x^n-a^n}{x-a}=na^{n-1}.$$

Proof. Multiplying out (the middle terms cancel in pairs) shows that $$\begin{aligned}x^n-a^n=(x-a)\big(&x^{n-1}+x^{n-2}a+x^{n-3}a^2\\&+\dots+xa^{n-2}+a^{n-1}\big).\end{aligned}$$ So for $x\ne a$ we may cancel $x-a$: $$\dfrac{x^n-a^n}{x-a}=x^{n-1}+x^{n-2}a+\dots+a^{n-1},$$ a polynomial with exactly $n$ terms. By Theorem 12.3 its limit as $x\to a$ is its value at $x=a$, where every term equals $a^{n-1}$. Hence the limit is $n\,a^{n-1}$. $\blacksquare$

Remark. The same formula holds for every rational number $n$ (for example $n=\tfrac12$ or $n=-1$) provided $a\gt0$. We do not prove this here, but you can check special cases: for $n=\tfrac12$ it says $\tfrac{\sqrt x-\sqrt a}{x-a}\to\tfrac{1}{2\sqrt a}$, which rationalising confirms, since the fraction equals $\tfrac{1}{\sqrt x+\sqrt a}$.

โœ๏ธ Example 4 โ€” Using the standard limit
Evaluate (a) $\displaystyle\lim_{x\to2}\dfrac{x^5-32}{x-2}$ (b) $\displaystyle\lim_{x\to1}\dfrac{x^{15}-1}{x^{10}-1}$ (c) $\displaystyle\lim_{x\to0}\dfrac{(1+x)^6-1}{x}$ (d) $\displaystyle\lim_{x\to9}\dfrac{\sqrt x-3}{x-9}$.
(a) $32=2^5$, so this is the standard form with $n=5$, $a=2$: $5\cdot2^4=80$.
(b) Divide top and bottom by $x-1$: $\dfrac{x^{15}-1}{x^{10}-1}=\dfrac{(x^{15}-1^{15})/(x-1)}{(x^{10}-1^{10})/(x-1)}\to\dfrac{15\cdot1^{14}}{10\cdot1^{9}}=\dfrac32$ (quotient rule, denominator limit $10\ne0$).
(c) Put $y=1+x$; as $x\to0$, $y\to1$, and the expression is $\dfrac{y^6-1^6}{y-1}\to6\cdot1^5=6$.
(d) $\sqrt x-3=x^{1/2}-9^{1/2}$: by the remark with $n=\tfrac12$, $a=9$, the limit is $\tfrac12\cdot9^{-1/2}=\tfrac16$. Check by rationalising: the fraction is $\tfrac{1}{\sqrt x+3}\to\tfrac16$ โœ“.
Interactive 1 โ€” Inside the 0/0 of (xโฟ โˆ’ aโฟ)/(x โˆ’ a)

The curve is $y=\tfrac{x^n-a^n}{x-a}$. At $x=a$ it is $\tfrac00$, so there is a hole; everywhere else it equals the polynomial $x^{n-1}+x^{n-2}a+\dots+a^{n-1}$. Choose $n$ and $a$, then press Play to send the red and blue dots in towards the hole from both sides. The readouts compare their heights with $na^{n-1}$ and list the $n$ terms of the polynomial at $x=a$ โ€” each one is $a^{n-1}$.

31.2
$h$โ€”
value at $a-h$โ€”
value at $a+h$โ€”
$na^{n-1}$โ€”
the $n$ terms at $x=a$โ€”

The Sandwich Theorem and $\dfrac{\sin x}{x}$

Substituting $x=0$ in $\tfrac{\sin x}{x}$ gives $\tfrac00$ again, but now there is nothing to factorise. A calculator (in radian mode) suggests the answer: $\tfrac{\sin0.1}{0.1}\approx0.998334$, $\tfrac{\sin0.01}{0.01}\approx0.999983$, and since $\tfrac{\sin(-x)}{-x}=\tfrac{\sin x}{x}$ the values for negative $x$ are the same. To prove that the limit is $1$ we trap $\tfrac{\sin x}{x}$ between two simpler functions that both tend to $1$.

๐Ÿ“ Theorem 12.5 โ€” The sandwich theorem

Let $f$, $g$ and $h$ be defined on an open interval containing $a$ (except possibly at $a$), with $$f(x)\le g(x)\le h(x)$$ for every $x\ne a$ in that interval. If $\displaystyle\lim_{x\to a}f(x)=\lim_{x\to a}h(x)=l$, then $\displaystyle\lim_{x\to a}g(x)=l$ as well.

Why. Near $a$ both $f(x)$ and $h(x)$ are within any chosen small distance of $l$; and $g(x)$, lying between them, can only be closer to $l$ than the further of the two. So $g(x)$ is also as close to $l$ as we like. $\blacksquare$

๐Ÿ“ Theorem 12.6 โ€” An inequality for $\dfrac{\sin x}{x}$

For every $x$ (in radians) with $0\lt|x|\lt\tfrac\pi2$, $$\cos x\lt\dfrac{\sin x}{x}\lt1.$$

Proof. First let $0\lt x\lt\tfrac\pi2$. In the unit circle with centre $O$ (Interactive 2 draws the figure), let $A=(1,0)$, let $B=(\cos x,\sin x)$ be the point at angle $x$, and let the tangent to the circle at $A$ meet the line $OB$ at $C=(1,\tan x)$. The region bounded by triangle $OAB$ lies inside the sector $OAB$, which lies inside triangle $OAC$. Comparing areas (base $OA=1$; a sector of angle $x$ in a circle of radius $1$ has area $\tfrac12x$):

$$\tfrac12\cdot1\cdot\sin x\ \lt\ \tfrac12x\ \lt\ \tfrac12\cdot1\cdot\tan x.$$

Multiply by $\tfrac{2}{\sin x}$, which is positive: $1\lt\dfrac{x}{\sin x}\lt\dfrac1{\cos x}$. Taking reciprocals of these positive numbers reverses the inequalities: $\cos x\lt\dfrac{\sin x}{x}\lt1$.

If $-\tfrac\pi2\lt x\lt0$, then $-x$ lies in $(0,\tfrac\pi2)$, and since $\cos(-x)=\cos x$ and $\tfrac{\sin(-x)}{-x}=\tfrac{\sin x}{x}$, the same inequality holds. $\blacksquare$

To finish we need $\cos x\to1$ as $x\to0$. The left half of the area inequality says $0\lt\sin x\lt x$ for $0\lt x\lt\tfrac\pi2$, so $|\sin x|\le|x|$ near $0$, and the sandwich theorem (with $-|x|\le\sin x\le|x|$) gives $\displaystyle\lim_{x\to0}\sin x=0$. Then $\cos x=1-2\sin^2\tfrac x2\to1-0=1$ (Section 3.4).

๐Ÿ“ Theorem 12.7 โ€” The two standard trigonometric limits

With $x$ measured in radians, $$\text{(a)}\ \ \lim_{x\to0}\dfrac{\sin x}{x}=1$$ $$\text{(b)}\ \ \lim_{x\to0}\dfrac{1-\cos x}{x}=0.$$

Proof. (a) By Theorem 12.6, $\cos x\lt\tfrac{\sin x}{x}\lt1$ for $0\lt|x|\lt\tfrac\pi2$. As $x\to0$ the left side tends to $1$ and the right side is $1$, so by the sandwich theorem $\tfrac{\sin x}{x}\to1$.

(b) Using $1-\cos x=2\sin^2\tfrac x2$: $$\dfrac{1-\cos x}{x}=\dfrac{2\sin^2\frac x2}{x}=\dfrac{\sin\frac x2}{\frac x2}\cdot\sin\dfrac x2.$$ As $x\to0$, $\tfrac x2\to0$, so the first factor tends to $1$ by (a), and the second to $\sin0=0$. By the product rule the limit is $1\cdot0=0$. $\blacksquare$

Three useful consequences follow at once, each by rewriting the expression in terms of $\tfrac{\sin\square}{\square}$ with the same "box" on top and bottom:

Interactive 2 โ€” The Squeeze on (sin x)/x

Left: the unit circle with the triangle $OAB$ (area $\tfrac12\sin x$), the sector $OAB$ (area $\tfrac12x$) and the triangle $OAC$ (area $\tfrac12\tan x$), nested inside one another. Right: the graphs of $\cos x$ (lower jaw), $\tfrac{\sin x}{x}$ (the filling) and $1$ (upper jaw). Press Play to shrink the angle $x$ and watch the jaws close on $1$.

1
areas in the unit circle
$\cos x\lt\frac{\sin x}{x}\lt1$
$\tfrac12\sin x$โ€”
$\tfrac12x$โ€”
$\tfrac12\tan x$โ€”
$\cos x$โ€”
$\tfrac{\sin x}{x}$โ€”
โœ๏ธ Example 5 โ€” Matching the box
Evaluate (a) $\displaystyle\lim_{x\to0}\dfrac{\sin4x}{x}$ (b) $\displaystyle\lim_{x\to0}\dfrac{\sin3x}{\sin5x}$ (c) $\displaystyle\lim_{x\to0}\dfrac{\tan2x}{x}$ (d) $\displaystyle\lim_{x\to0}\dfrac{1-\cos2x}{x^2}$.
(a) $\dfrac{\sin4x}{x}=4\cdot\dfrac{\sin4x}{4x}\to4\cdot1=4$.
(b) $\dfrac{\sin3x}{\sin5x}=\dfrac{\frac{\sin3x}{3x}\cdot3x}{\frac{\sin5x}{5x}\cdot5x}=\dfrac35\cdot\dfrac{\frac{\sin3x}{3x}}{\frac{\sin5x}{5x}}\to\dfrac35\cdot\dfrac11=\dfrac35$.
(c) $\dfrac{\tan2x}{x}=2\cdot\dfrac{\sin2x}{2x}\cdot\dfrac1{\cos2x}\to2\cdot1\cdot1=2$.
(d) $1-\cos2x=2\sin^2x$, so the expression is $2\left(\dfrac{\sin x}{x}\right)^2\to2$.
Check At $x=0.001$: (b) $\tfrac{\sin0.003}{\sin0.005}\approx0.6000016$; (d) $\tfrac{1-\cos0.002}{10^{-6}}\approx1.9999993$ โœ“.
โœ๏ธ Example 6 โ€” Identities first, then the standard limits
Evaluate (a) $\displaystyle\lim_{x\to0}\dfrac{1-\cos x}{x\sin x}$ (b) $\displaystyle\lim_{x\to\pi}\dfrac{\sin x}{\pi-x}$ (c) $\displaystyle\lim_{x\to0}\dfrac{\cos3x-\cos x}{x^2}$.
(a) $\dfrac{2\sin^2\frac x2}{x\cdot2\sin\frac x2\cos\frac x2}=\dfrac{\sin\frac x2}{x\cos\frac x2}$ $=\dfrac12\cdot\dfrac{\sin\frac x2}{\frac x2}\cdot\dfrac1{\cos\frac x2}\to\dfrac12$.
(b) The limit is not at $0$, so shift it there: put $y=\pi-x$, so $y\to0$ as $x\to\pi$, and $\sin x=\sin(\pi-y)=\sin y$. The limit is $\displaystyle\lim_{y\to0}\dfrac{\sin y}{y}=1$.
(c) By the sum-to-product formula, $\cos3x-\cos x=-2\sin2x\sin x$. So the expression is $-2\cdot\dfrac{\sin2x}{x}\cdot\dfrac{\sin x}{x}\to-2\cdot2\cdot1=-4$.
Check (c) at $x=0.001$: $\cos0.003-\cos0.001\approx-0.0000040000$, divided by $10^{-6}$ gives $\approx-4.0000$ โœ“.
โš ๏ธ Common mistakes with these limits

1. Cancelling $x$ in $\tfrac{\sin x}{x}$. $\sin x$ is not "$\sin$ times $x$"; there is nothing to cancel. Likewise $\tfrac{\sin4x}{x}$ is not $\sin4$ and $\tfrac{\sin3x}{\sin5x}$ is not $\tfrac{3}{5}$ "by cancelling $\sin$" โ€” the answer happens to be $\tfrac35$, but only because of the standard limit.

2. Forgetting to match the box. $\tfrac{\sin4x}{x}\to1$ is wrong: the angle and the denominator must be the same, so write $4\cdot\tfrac{\sin4x}{4x}$.

3. Degrees. Theorem 12.7 needs radians: the sector area $\tfrac12x$ is a radian formula. In degrees, $\displaystyle\lim_{x\to0}\tfrac{\sin x^\circ}{x}=\tfrac{\pi}{180}$, because $x^\circ=\tfrac{\pi x}{180}$ radians.

4. Using $\tfrac{\sin x}{x}\to1$ when $x$ does not tend to $0$. $\displaystyle\lim_{x\to\pi}\tfrac{\sin x}{x}=\tfrac{0}{\pi}=0$ by plain substitution. The standard limit is about the box tending to $0$.

5. Not checking for $\tfrac00$ first. Always substitute first; factorising $\tfrac{x^2-1}{x+1}$ at $x=1$ is unnecessary ($\tfrac02=0$), and "cancelling" in $\tfrac{x+1}{x-1}$ at $x=1$ is impossible ($\tfrac20$: no finite limit).

6. Cancelling without $x\ne a$. $\tfrac{(x-2)(x+2)}{x-2}=x+2$ is true only for $x\ne2$. That is exactly the case a limit looks at, so the step is valid inside a limit โ€” but $\tfrac{x^2-4}{x-2}$ and $x+2$ are still different functions.

๐Ÿ“˜ In the CBSE content list, but not taught in the NCERT textbook (also JEE Main and JEE Advanced)

The CBSE 2026-27 syllabus lists limits of exponential and logarithmic functions, but the NCERT textbook has no section on them, so board questions on them are rare; JEE Main and JEE Advanced both use two more standard limits, for the exponential and logarithm functions you will meet properly in Class 12.

The number $e\approx2.71828$ is the base for which $$\lim_{x\to0}\dfrac{e^x-1}{x}=1,$$ and then $$\lim_{x\to0}\dfrac{\log(1+x)}{x}=1,$$ where $\log$ means the natural logarithm, to base $e$. A table supports both: at $x=0.001$, $\tfrac{e^{0.001}-1}{0.001}\approx1.0005$ and $\tfrac{\log1.001}{0.001}\approx0.9995$. They are used exactly like $\tfrac{\sin x}{x}$: match the box, then the box tends to $0$.

Worked example. Find $\displaystyle\lim_{x\to0}\dfrac{e^{3x}-1}{\log(1+2x)}$.
Solution. $\dfrac{e^{3x}-1}{\log(1+2x)}=\dfrac{\frac{e^{3x}-1}{3x}\cdot3x}{\frac{\log(1+2x)}{2x}\cdot2x}$ $=\dfrac32\cdot\dfrac{\frac{e^{3x}-1}{3x}}{\frac{\log(1+2x)}{2x}}$ $\to\dfrac32\cdot\dfrac11=\dfrac32$. Check: at $x=0.0001$ the value is $\tfrac{0.00030004500}{0.00019998000}\approx1.50038$ โœ“.

Exercises

I
Polynomials and rational functions: substitute, factorise.
  • 1 $\displaystyle\lim_{x\to3}(x^2+2x-1)$
  • 2 $\displaystyle\lim_{x\to-2}(x^3-4x+1)$
  • 3 $\displaystyle\lim_{x\to1}\dfrac{x^2+1}{x+3}$
  • 4 $\displaystyle\lim_{x\to4}\dfrac{x^2-16}{x-4}$
  • 5 $\displaystyle\lim_{x\to2}\dfrac{x^2-3x+2}{x^2-4}$
  • 6 $\displaystyle\lim_{x\to-1}\dfrac{x^3+1}{x+1}$
  • 7 $\displaystyle\lim_{x\to3}\dfrac{x^3-27}{x^2-9}$
  • 8 $\displaystyle\lim_{x\to2}\dfrac{x^3-3x^2+4}{x^2-4x+4}$
  • 9 $\displaystyle\lim_{x\to0}\dfrac{ax+bx^2}{cx+dx^2}$, where $c\ne0$
  • 10 $\displaystyle\lim_{x\to1}\left(\dfrac{1}{x-1}-\dfrac{2}{x^2-1}\right)$
1 $9+6-1=14$.
2 $-8+8+1=1$.
3 $\tfrac24=\tfrac12$.
4 $\tfrac00$; $\tfrac{(x-4)(x+4)}{x-4}=x+4\to8$.
5 $\tfrac00$; $\tfrac{(x-1)(x-2)}{(x-2)(x+2)}=\tfrac{x-1}{x+2}\to\tfrac14$.
6 $\tfrac00$; $x^3+1=(x+1)(x^2-x+1)$, so the limit is $1+1+1=3$.
7 $\tfrac00$; $\tfrac{(x-3)(x^2+3x+9)}{(x-3)(x+3)}\to\tfrac{27}{6}=\tfrac92$.
8 $\tfrac00$ at $x=2$. Dividing the numerator by $x-2$ twice gives $x^3-3x^2+4=(x-2)^2(x+1)$, and the denominator is $(x-2)^2$. The fraction is $x+1$ for $x\ne2$, so the limit is $3$.
9 Cancel $x$: $\dfrac{a+bx}{c+dx}\to\dfrac ac$.
10 Combine first: $\dfrac{(x+1)-2}{x^2-1}=\dfrac{x-1}{(x-1)(x+1)}=\dfrac{1}{x+1}\to\dfrac12$. (Each fraction separately has no finite limit, so Theorem 12.2 cannot be applied to them one at a time.)
II
Rationalising and the standard limit $\tfrac{x^n-a^n}{x-a}$.
  • 1 $\displaystyle\lim_{x\to0}\dfrac{\sqrt{9+x}-3}{x}$
  • 2 $\displaystyle\lim_{x\to2}\dfrac{\sqrt{x+2}-2}{x-2}$
  • 3 $\displaystyle\lim_{x\to0}\dfrac{x}{\sqrt{1+x}-\sqrt{1-x}}$
  • 4 $\displaystyle\lim_{x\to3}\dfrac{x^4-81}{x-3}$
  • 5 $\displaystyle\lim_{x\to1}\dfrac{x^7-1}{x-1}$
  • 6 $\displaystyle\lim_{x\to1}\dfrac{x^9-1}{x^6-1}$
  • 7 $\displaystyle\lim_{x\to-2}\dfrac{x^5+32}{x+2}$
  • 8 $\displaystyle\lim_{x\to0}\dfrac{(1+x)^4-1}{x}$
  • 9 Find the positive integer $n$ for which $\displaystyle\lim_{x\to2}\dfrac{x^n-2^n}{x-2}=80$.
1 Multiply by $\sqrt{9+x}+3$: $\dfrac{1}{\sqrt{9+x}+3}\to\dfrac16$.
2 Multiply by $\sqrt{x+2}+2$: $\dfrac{(x+2)-4}{(x-2)(\sqrt{x+2}+2)}=\dfrac1{\sqrt{x+2}+2}\to\dfrac14$.
3 Multiply by $\sqrt{1+x}+\sqrt{1-x}$: the denominator becomes $(1+x)-(1-x)=2x$, so the fraction is $\dfrac{\sqrt{1+x}+\sqrt{1-x}}{2}\to1$.
4 $81=3^4$: $4\cdot3^3=108$.
5 $7\cdot1^6=7$.
6 Divide top and bottom by $x-1$: $\tfrac{9}{6}=\tfrac32$.
7 $-32=(-2)^5$ and $x+2=x-(-2)$: $5(-2)^4=80$.
8 With $y=1+x\to1$: $\dfrac{y^4-1}{y-1}\to4$. (Or expand: $\tfrac{4x+6x^2+4x^3+x^4}{x}\to4$.)
9 $n\cdot2^{n-1}=80$. Trying $n=1,2,3,4,5$: $1,4,12,32,80$; so $n=5$ (the left side increases with $n$, so the answer is unique).
III
Trigonometric limits ($x$ in radians unless stated).
  • 1 $\displaystyle\lim_{x\to0}\dfrac{\sin5x}{x}$
  • 2 $\displaystyle\lim_{x\to0}\dfrac{\sin2x}{\sin7x}$
  • 3 $\displaystyle\lim_{x\to0}\dfrac{\tan4x}{3x}$
  • 4 $\displaystyle\lim_{x\to0}\dfrac{1-\cos4x}{x^2}$
  • 5 $\displaystyle\lim_{x\to0}x\cot x$
  • 6 $\displaystyle\lim_{x\to0}\dfrac{\sin ax+bx}{ax+\sin bx}$, where $a$, $b$ and $a+b$ are non-zero
  • 7 $\displaystyle\lim_{x\to0}\dfrac{\sin^23x}{x^2}$
  • 8 $\displaystyle\lim_{x\to0}\dfrac{\cos2x-1}{\cos x-1}$
  • 9 $\displaystyle\lim_{x\to\pi/2}\dfrac{\cos x}{\frac\pi2-x}$
  • 10 $\displaystyle\lim_{x\to0}\dfrac{\tan x-\sin x}{x^3}$
  • 11 $\displaystyle\lim_{x\to0}\dfrac{\sin x^\circ}{x}$, where $x^\circ$ means $x$ degrees
Check: values of the Card III expressions near the limit point
1 $5\cdot\tfrac{\sin5x}{5x}\to5$.
2 $\tfrac27\cdot\dfrac{\sin2x/(2x)}{\sin7x/(7x)}\to\tfrac27$.
3 $\tfrac43\cdot\tfrac{\sin4x}{4x}\cdot\tfrac{1}{\cos4x}\to\tfrac43$.
4 $1-\cos4x=2\sin^22x$: $2\cdot4\left(\tfrac{\sin2x}{2x}\right)^2\to8$.
5 $x\cot x=\dfrac{x}{\sin x}\cdot\cos x\to1\cdot1=1$.
6 Divide top and bottom by $x$: $\dfrac{a\frac{\sin ax}{ax}+b}{a+b\frac{\sin bx}{bx}}\to\dfrac{a+b}{a+b}=1$.
7 $9\left(\tfrac{\sin3x}{3x}\right)^2\to9$.
8 $\dfrac{-2\sin^2x}{-2\sin^2\frac x2}=\dfrac{\left(\frac{\sin x}{x}\right)^2x^2}{\left(\frac{\sin(x/2)}{x/2}\right)^2\frac{x^2}{4}}\to4$.
9 Put $y=\tfrac\pi2-x\to0$: $\cos x=\cos\left(\tfrac\pi2-y\right)=\sin y$, so the limit is $\displaystyle\lim_{y\to0}\tfrac{\sin y}{y}=1$.
10 $\tan x-\sin x=\tan x(1-\cos x)$, so the expression is $\dfrac{\tan x}{x}\cdot\dfrac{1-\cos x}{x^2}\to1\cdot\tfrac12=\tfrac12$.
11 $\sin x^\circ=\sin\tfrac{\pi x}{180}$, so the limit is $\tfrac{\pi}{180}\approx0.01745$.
IV
Applications.
  • 1 A regular polygon with $n$ sides is inscribed in a circle of radius $r$. Show that its area is $A_n=\tfrac n2r^2\sin\tfrac{2\pi}{n}$. Find $A_6$, $A_{12}$ and $A_{96}$ for $r=1$, and use $\tfrac{\sin\theta}{\theta}\to1$ to show that $A_n$ approaches $\pi r^2$ as $n$ grows.
  • 2 Engineers often use the small-angle approximation $\sin\theta\approx\theta$. Find the percentage error of this approximation, $\tfrac{\theta-\sin\theta}{\sin\theta}\times100\%$, for $\theta=0.1$ rad and for $\theta=5^\circ$.
  • 3 A railway track follows a circular arc of radius $500\ \text{m}$ through an angle of $0.02$ rad. Find the length of the arc and of the straight chord joining its ends, and the difference in millimetres. What is $\displaystyle\lim_{\theta\to0}\tfrac{\text{chord}}{\text{arc}}$ for a circle of radius $r$?
1 Joining the centre to the vertices cuts the polygon into $n$ triangles with two sides $r$ and angle $\tfrac{2\pi}n$ between them, each of area $\tfrac12r^2\sin\tfrac{2\pi}n$. With $r=1$: $A_6=3\sin\tfrac\pi3\approx2.598$, $A_{12}=6\sin\tfrac\pi6=3$, $A_{96}=48\sin\tfrac{\pi}{48}\approx3.1394$. Writing $\theta=\tfrac{2\pi}n$: $A_n=\pi r^2\cdot\dfrac{\sin\theta}{\theta}$, and $\theta\to0$ as $n$ grows, so $A_n\to\pi r^2$.
2 $\theta=0.1$: $\sin0.1\approx0.0998334$, error $\approx0.167\%$. $\theta=5^\circ\approx0.0872665$ rad: $\sin\theta\approx0.0871557$, error $\approx0.127\%$. Both are tiny, which is why the approximation is so popular.
3 Arc $=r\theta=500\times0.02=10\ \text{m}$. Chord $=2r\sin\tfrac\theta2=1000\sin0.01\approx9.999833\ \text{m}$. The difference is about $0.000167\ \text{m}\approx0.17\ \text{mm}$. In general $\dfrac{\text{chord}}{\text{arc}}=\dfrac{2r\sin(\theta/2)}{r\theta}=\dfrac{\sin(\theta/2)}{\theta/2}\to1$.
๐ŸŽฏ Exam corner Board ยท 1 markBoard ยท 2 marksBoard ยท 3 marksJEE MainJEE Advanced

How this is examined: almost every Board paper has a $\tfrac00$ limit to factorise or rationalise (2 marks), one using $\tfrac{x^n-a^n}{x-a}$, and a trigonometric limit needing an identity before $\tfrac{\sin x}{x}\to1$ (2โ€“3 marks). JEE Main combines several standard limits in one expression; JEE Advanced asks for higher-order behaviour such as $\tfrac{1-\cos x}{x^2}\to\tfrac12$.

E1. Board ยท 1 mark $\displaystyle\lim_{x\to0}\dfrac{\sin3x}{\tan2x}$ is (A) $\tfrac32$   (B) $\tfrac23$   (C) $1$   (D) $0$

Idea $\dfrac{\sin3x}{\tan2x}=\dfrac32\cdot\dfrac{\sin3x/(3x)}{\tan2x/(2x)}\to\dfrac32\cdot\dfrac11$.
Trap (B) turns the ratio upside down; (C) "cancels" the trigonometric functions and the $x$'s; (D) substitutes $\sin0=0$ in the numerator but forgets the denominator is $0$ too.
Answer (A).

E2. Board ยท 2 marks Evaluate $\displaystyle\lim_{x\to2}\dfrac{x^3-8}{x^2-3x+2}$.

Factorise At $x=2$ the form is $\tfrac00$. $x^3-8=(x-2)(x^2+2x+4)$ and $x^2-3x+2=(x-1)(x-2)$. (1 mark)
Answer $\displaystyle\lim_{x\to2}\dfrac{x^2+2x+4}{x-1}=\dfrac{12}{1}=12$. (1 mark)

E3. Board ยท 2 marks Find all real $k$ for which $\displaystyle\lim_{x\to1}\dfrac{x^6-1}{x-1}=\lim_{x\to k}\dfrac{x^3-k^3}{x-k}$.

Both sides By Theorem 12.4 the left side is $6\cdot1^5=6$ and the right side is $3k^2$. (1 mark)
Answer $3k^2=6$, so $k=\pm\sqrt2$. (1 mark)

E4. Board ยท 3 marks Prove that $\displaystyle\lim_{x\to0}\dfrac{\cos ax-\cos bx}{x^2}=\dfrac{b^2-a^2}{2}$ for any non-zero constants $a$, $b$, and hence find $\displaystyle\lim_{x\to0}\dfrac{\cos x-\cos3x}{x^2}$.

Identity $\cos ax-\cos bx$ $=-2\sin\dfrac{(a+b)x}{2}\sin\dfrac{(a-b)x}{2}$. (1 mark)
Limit If $a\ne\pm b$: $\dfrac{\cos ax-\cos bx}{x^2}$ $=-2\cdot\dfrac{\sin\frac{(a+b)x}{2}}{x}\cdot\dfrac{\sin\frac{(a-b)x}{2}}{x}$ $\to-2\cdot\dfrac{a+b}{2}\cdot\dfrac{a-b}{2}$ $=\dfrac{b^2-a^2}{2}$. (If $a=\pm b$ the expression is $0$ and so is $\tfrac{b^2-a^2}2$.) (1 mark)
Answer $a=1$, $b=3$: $\dfrac{9-1}{2}=4$. Check: at $x=0.001$, $\tfrac{\cos0.001-\cos0.003}{10^{-6}}\approx4.0000$ โœ“. (1 mark)

E5. JEE Main $\displaystyle\lim_{x\to0}\dfrac{(1-\cos4x)(2+\cos x)}{x\tan2x}$ equals (A) $12$   (B) $6$   (C) $4$   (D) $24$

Method $1-\cos4x=2\sin^22x$. The expression is $\dfrac{2\sin^22x}{x\tan2x}(2+\cos x)$ $=2\cdot\dfrac{\sin2x}{x}\cdot\dfrac{\sin2x}{\tan2x}\cdot(2+\cos x)$ $=2\cdot\dfrac{\sin2x}{x}\cdot\cos2x\cdot(2+\cos x)$.
Limit $\to2\cdot2\cdot1\cdot3=12$.
Trap (C) forgets the factor $2+\cos x\to3$; (D) uses $1-\cos4x\approx4x^2\cdot2$ wrongly as $16x^2$ instead of $8x^2$; (B) loses the $2$ in $2\sin^22x$.
โฑ Fast route Near $0$: $1-\cos4x\approx\tfrac{(4x)^2}{2}=8x^2$, $\tan2x\approx2x$, $2+\cos x\approx3$: $\tfrac{8x^2\cdot3}{2x^2}=12$.
Answer (A).

E6. JEE Advanced Numerical answer. Find $\displaystyle\lim_{x\to0}\dfrac{\sin2x-2\sin x}{x^3}$.

Factor $\sin2x-2\sin x=2\sin x\cos x-2\sin x$ $=-2\sin x(1-\cos x)$.
Limit $\dfrac{-2\sin x(1-\cos x)}{x^3}$ $=-2\cdot\dfrac{\sin x}{x}\cdot\dfrac{1-\cos x}{x^2}$ $\to-2\cdot1\cdot\dfrac12=-1$.
Check At $x=0.01$: $\sin0.02-2\sin0.01\approx-0.00000099998$, divided by $10^{-6}$: $\approx-0.99998$ โœ“.
Answer $\mathbf{-1}$.
๐Ÿ”ฌ Researchโ€“Teaching NexusMini project ยท about 60 minutes

Research question. For which angles is the small-angle approximation $\sin\theta\approx\theta$ good enough for a pendulum clock, a camera's field of view or a surveyor's measurement โ€” and how does its error grow?

Why it matters. The limit $\tfrac{\sin\theta}{\theta}\to1$ is the reason physicists can replace $\sin\theta$ by $\theta$ in the pendulum equation, astronomers can treat the tiny angle a distant star makes as its width divided by its distance, and optical engineers can design lenses with simple formulae. Every one of them needs to know when the approximation stops being safe.

What to do.

  1. In a spreadsheet, list $\theta$ from $1^\circ$ to $45^\circ$. Convert each to radians and compute $\sin\theta$, the ratio $\tfrac{\sin\theta}{\theta}$ and the percentage error $\tfrac{\theta-\sin\theta}{\sin\theta}\times100\%$.
  2. Plot the percentage error against $\theta$. Find the largest angle for which the error is below $1\%$ and below $0.1\%$.
  3. Divide each error by $\theta^2$ (in radians). What number does this approach as $\theta\to0$? Explain it using $\tfrac{1-\cos x}{x^2}\to\tfrac12$ as a guide (the true value is $\tfrac16$).
  4. Make a simple pendulum (a nut on a $1\ \text{m}$ thread). Time $10$ swings for release angles of $5^\circ$, $15^\circ$, $30^\circ$ and $45^\circ$, using a protractor. Does the period change once the approximation becomes poor?

Tools: A spreadsheet, a protractor, thread and a small weight, a phone stopwatch

Hand in: Your table and error graph, the two threshold angles, your value for the limit in step 3, and the pendulum timings with a sentence linking them to the error graph.

Going further. Repeat steps 1โ€“2 for $\tan\theta\approx\theta$ and $\cos\theta\approx1-\tfrac{\theta^2}{2}$. Which approximation fails first?

Instructor note: the error of $\sin\theta\approx\theta$ is below $1\%$ up to about $14^\circ$ and below $0.1\%$ up to about $4.4^\circ$; error$/\theta^2\to\tfrac16\approx0.1667$ (as a fraction, not a percentage). A $1\ \text{m}$ pendulum has period about $2.01\ \text{s}$ at small angles; at $45^\circ$ it is about $4\%$ longer, at $15^\circ$ only about $0.4\%$ โ€” usually hard to detect with hand timing, which is itself a useful discussion.

๐Ÿงช Self-check quiz

0/5 correct

Q1. $\displaystyle\lim_{x\to2}\dfrac{x^2+3x}{x+2}$ is

Q2. $\displaystyle\lim_{x\to1}\dfrac{x^3-1}{x-1}$ is

Q3. $\displaystyle\lim_{x\to0}\dfrac{\sin6x}{2x}$ is

Q4. $\displaystyle\lim_{x\to0}\dfrac{1-\cos x}{x}$ is

Q5. The proof that $\tfrac{\sin x}{x}\to1$ traps it between

โ† 12.1 Intuitive Idea of Limits12.3 The Derivative from First Principles โ†’