NCERT Exemplar Solutions Class 11 Maths Limits and Derivatives — Short Answer Type Q1–14

NCERT Exemplar Problems, Class 11 Mathematics — Limits and Derivatives, Short Answer Type questions 1 to 14. All 14 questions solved.

These fourteen questions are all limits of algebraic expressions, and almost every one is a $\frac00$ form. Two tools do all the work: cancel a common factor that causes the zero (by factorising, or by rationalising a square root), or use the standard limit $\displaystyle\lim_{x\to a}\frac{x^n-a^n}{x-a}=na^{n-1}$, which holds for any rational $n$.

Two questions are misprinted in the book, Questions 5 and 9. Each is shown as printed and as it was evidently meant; the explanation is given where it occurs.

Key insight. A limit as $x\to a$ never asks what happens at $x=a$. When direct substitution gives $\frac00$, the numerator and denominator share a factor $(x-a)$; remove it, then substitute. Rationalising and the formula for $\frac{x^n-a^n}{x-a}$ are just two ways of exposing that factor.

Evaluate:

Question 1

$\displaystyle\lim_{x\to3}\frac{x^2-9}{x-3}$

Solution. Direct substitution gives $\frac00$. The numerator is a difference of squares, so the factor $x-3$ is sitting there waiting to be cancelled. Since $x\ne3$ in a limit, the cancellation is legitimate.

$$\lim_{x\to3}\frac{(x-3)(x+3)}{x-3}=\lim_{x\to3}(x+3)=6$$

Question 2

$\displaystyle\lim_{x\to\frac12}\frac{4x^2-1}{2x-1}$

Solution. At $x=\frac12$ both parts vanish. $4x^2-1=(2x-1)(2x+1)$ is again a difference of squares.

$$\lim_{x\to\frac12}\frac{(2x-1)(2x+1)}{2x-1}=\lim_{x\to\frac12}(2x+1)=2$$

Question 3

$\displaystyle\lim_{h\to0}\frac{\sqrt{x+h}-\sqrt x}{h}$

Solution. The variable here is $h$; $x$ is a constant. The zero in the numerator comes from a difference of square roots, so multiply top and bottom by the conjugate $\sqrt{x+h}+\sqrt x$, which turns the numerator into $(x+h)-x=h$.

$$\lim_{h\to0}\frac{(x+h)-x}{h\left(\sqrt{x+h}+\sqrt x\right)}=\lim_{h\to0}\frac{1}{\sqrt{x+h}+\sqrt x}=\frac{1}{2\sqrt x}$$

This is just the derivative of $\sqrt x$ from first principles.

Question 4

$\displaystyle\lim_{x\to0}\frac{(x+2)^{1/3}-2^{1/3}}{x}$

Solution. Put $y=x+2$, so that $y\to2$ as $x\to0$ and $x=y-2$. The expression becomes $\dfrac{y^{1/3}-2^{1/3}}{y-2}$, exactly the form $\dfrac{y^n-a^n}{y-a}$ with $n=\frac13$ and $a=2$. Its limit is $na^{n-1}$.

$$\lim_{y\to2}\frac{y^{1/3}-2^{1/3}}{y-2}=\frac13\cdot2^{\frac13-1}=\frac13\cdot2^{-2/3}$$

Question 5

$\displaystyle\lim_{x\to1}\frac{(1+x)^6-1}{(1+x)^2-1}$

Solution. The book prints $x\to1$, and the key’s answer $3$ belongs to $x\to0$, so the question is misprinted. As printed, the denominator at $x=1$ is $2^2-1=3$, which is not zero, so this is not a $\frac00$ form at all and direct substitution is valid:

$$\frac{2^6-1}{2^2-1}=\frac{63}{3}=21.$$

If the limit was meant to be $x\to0$, both parts vanish, and with $u=(1+x)^2$ the expression is $\dfrac{u^3-1}{u-1}=u^2+u+1\to1+1+1=3$, which is the key’s value.

Question 6

$\displaystyle\lim_{x\to a}\frac{(2+x)^{5/2}-(a+2)^{5/2}}{x-a}$

Solution. Substitute $y=2+x$ and $b=a+2$. Then $x-a=y-b$ and $y\to b$, so the expression is $\dfrac{y^{5/2}-b^{5/2}}{y-b}$ with limit $\frac52b^{3/2}$.

$$\lim=\frac52(a+2)^{3/2}$$

Question 7

$\displaystyle\lim_{x\to1}\frac{x^4-\sqrt x}{\sqrt x-1}$

Solution. The root makes this awkward in $x$, so write $u=\sqrt x$, with $u\to1$. Then $x^4=u^8$ and the expression is $\dfrac{u^8-u}{u-1}=\dfrac{u(u^7-1)}{u-1}$. Now use the standard limit for $\dfrac{u^7-1}{u-1}$, which tends to $7$.

$$\lim_{u\to1}u\cdot\frac{u^7-1}{u-1}=1\cdot7=7$$

Question 8

$\displaystyle\lim_{x\to2}\frac{x^2-4}{\sqrt{3x-2}-\sqrt{x+2}}$

Solution. The roots sit in the denominator, so rationalise there. Multiplying by $\sqrt{3x-2}+\sqrt{x+2}$ turns the denominator into $(3x-2)-(x+2)=2x-4=2(x-2)$, which contains the troublesome factor.

$$\frac{(x-2)(x+2)\left(\sqrt{3x-2}+\sqrt{x+2}\right)}{2(x-2)}=\frac{(x+2)\left(\sqrt{3x-2}+\sqrt{x+2}\right)}{2}$$

At $x=2$: $\dfrac{4\,(2+2)}{2}=8$.

Question 9

$\displaystyle\lim_{x\to\sqrt2}\frac{x^4-4}{x^2+3\sqrt{2x}-8}$ (as printed)

Solution. As printed, the radical covers $2x$, and then the limit is $0$: the numerator tends to $4-4=0$, but the denominator tends to $2+3\cdot2^{3/4}-8\ne0$, so the quotient tends to $0$ and there is no $\frac00$ at all. The key’s $\frac85$ is what you get when the term is $3\sqrt2\,x$, which is plainly the intended question, because then the denominator also vanishes at $\sqrt2$.

The intended question is $\displaystyle\lim_{x\to\sqrt2}\frac{x^4-4}{x^2+3\sqrt2\,x-8}$. Factorise: $x^4-4=(x^2-2)(x^2+2)=(x-\sqrt2)(x+\sqrt2)(x^2+2)$, and $x^2+3\sqrt2x-8=(x-\sqrt2)(x+4\sqrt2)$ (check: the middle term is $4\sqrt2x-\sqrt2x=3\sqrt2x$ and the constant is $-4\cdot2=-8$). Cancel $x-\sqrt2$:

$$\lim_{x\to\sqrt2}\frac{(x+\sqrt2)(x^2+2)}{x+4\sqrt2}=\frac{2\sqrt2\cdot4}{5\sqrt2}=\frac85$$

Question 10

$\displaystyle\lim_{x\to1}\frac{x^7-2x^5+1}{x^3-3x^2+2}$

Solution. Both polynomials equal zero at $x=1$, so both have the factor $x-1$. Dividing out:

$$x^7-2x^5+1=(x-1)(x^6+x^5-x^4-x^3-x^2-x-1),\qquad x^3-3x^2+2=(x-1)(x^2-2x-2).$$

After cancelling $x-1$ and putting $x=1$ the quotients are $\dfrac{1+1-1-1-1-1-1}{1-2-2}=\dfrac{-3}{-3}=1$. As a check, L’Hôpital’s rule agrees: $\dfrac{7-10}{3-6}=1$.

Question 11

$\displaystyle\lim_{x\to0}\frac{\sqrt{1+x^3}-\sqrt{1-x^3}}{x^2}$

Solution. Rationalise the numerator. The numerator becomes $(1+x^3)-(1-x^3)=2x^3$.

$$\lim_{x\to0}\frac{2x^3}{x^2\left(\sqrt{1+x^3}+\sqrt{1-x^3}\right)}=\lim_{x\to0}\frac{2x}{\sqrt{1+x^3}+\sqrt{1-x^3}}=\frac{0}{2}=0$$

The numerator is of order $x^3$ and the denominator of order $x^2$, so the quotient must vanish.

Question 12

$\displaystyle\lim_{x\to-3}\frac{x^3+27}{x^5+243}$

Solution. Here $27=(-(-3))^3$ and $243=3^5$, so $x^3+27=x^3-(-3)^3$ and $x^5+243=x^5-(-3)^5$. Divide the top and the bottom by $x-(-3)=x+3$ and use $\dfrac{x^n-a^n}{x-a}\to na^{n-1}$ with $a=-3$:

$$\lim=\frac{3(-3)^2}{5(-3)^4}=\frac{27}{405}=\frac1{15}$$

Question 13

$\displaystyle\lim_{x\to\frac12}\left(\frac{8x-3}{2x-1}-\frac{4x^2+1}{4x^2-1}\right)$

Solution. Each fraction blows up at $x=\frac12$ separately, so combine them over the common denominator $(2x-1)(2x+1)$ before doing anything else.

$$\frac{(8x-3)(2x+1)-(4x^2+1)}{(2x-1)(2x+1)}=\frac{16x^2+2x-3-4x^2-1}{(2x-1)(2x+1)}=\frac{12x^2+2x-4}{(2x-1)(2x+1)}$$

The numerator is $2(6x^2+x-2)=2(2x-1)(3x+2)$. Cancel $2x-1$:

$$\lim_{x\to\frac12}\frac{2(3x+2)}{2x+1}=\frac{2\cdot\frac72}{2}=\frac72$$

Question 14

Find $n$, if $\displaystyle\lim_{x\to2}\frac{x^n-2^n}{x-2}=80,\ n\in\mathbf N$.

Solution. The limit is the standard one with $a=2$, so it equals $n\cdot2^{n-1}$. We need $n\cdot2^{n-1}=80$. The left side increases with $n$, so there is at most one solution; trying $n=4$ gives $4\cdot8=32$ and $n=5$ gives $5\cdot16=80$.

Common mistakes

• Question 5 (and 9). Assuming every limit is a $\frac00$ form. Question 5 as printed is not: substituting $x=1$ gives $\frac{63}{3}$. Always substitute first; only if it gives $\frac00$ do you need to cancel.
• Question 3. Treating $x$ as the variable. The limit is in $h$; the answer $\frac1{2\sqrt x}$ must still contain $x$.
• Question 4. Using $na^{n-1}$ without first shifting the variable. The denominator is $x$, not $(x+2)-2$ written out, so students forget that $x+2$ is the one that tends to $2$.
• Question 8. Rationalising the numerator, which is already polynomial, instead of the denominator that holds the roots.
• Question 12. Writing $a=3$ instead of $a=-3$ in $\frac{x^n-a^n}{x-a}$, which loses the sign and turns $\frac1{15}$ into a wrong value. $x^3+27$ is $x^3-a^3$ with $a=-3$.
• Question 13. Taking the limit of each fraction separately: both are undefined at $x=\frac12$, so the difference has to be combined before the limit means anything.

Practise next

• The rest of this unit: Short Answer Type Q15–28 (trigonometric limits), Short and Long Answer Type Q29–46, Long Answer and Objective Type Q47–64 and Objective and Fill in the Blanks Q65–80.
• NCERT textbook: Exercise 12.1, Exercise 12.2 and the Miscellaneous Exercise.