NCERT Solutions: Class 11 Maths Chapter 12 - Limits and Derivatives

The function is continuous, so substitute $x=3$: $3+3=6$.

Substitute $x=\pi$ directly to get $\pi-\dfrac{22}{7}$.

Substitute $r=1$: $\pi(1)^2=\pi$.

Substitute $x=4$: $\dfrac{4(4)+3}{4-2}=\dfrac{19}{2}$.

Substitute $x=-1$: numerator $=1-1+1=1$ and denominator $=-2$. Hence the limit is $-\dfrac12$.

Expand $(1+x)^5=1+5x+10x^2+\cdots$. Then $\dfrac{(1+x)^5-1}{x}=5+10x+\cdots$, whose limit as $x\to0$ is $5$.

Factor: $3x^2-x-10=(3x+5)(x-2)$ and $x^2-4=(x-2)(x+2)$. Cancel $x-2$ and substitute $x=2$: $\dfrac{3(2)+5}{2+2}=\dfrac{11}{4}$.

Factor $x^4-81=(x-3)(x+3)(x^2+9)$ and $2x^2-5x-3=(2x+1)(x-3)$. Cancel $x-3$ and substitute $x=3$: $\dfrac{(6)(18)}{7}=\dfrac{108}{7}$.

Substitute $x=0$: $\dfrac{a(0)+b}{c(0)+1}=b$.

Let $t=z^{1/6}$. Then $z^{1/3}=t^2$ and $t\to1$. The limit becomes $\lim_{t\to1}\dfrac{t^2-1}{t-1}=\lim_{t\to1}(t+1)=2$.

Substitute $x=1$. The numerator is $a+b+c$ and the denominator is $c+b+a=a+b+c$. Since $a+b+c\ne0$, the limit is $1$.

Simplify the numerator: $\dfrac1x+\dfrac12=\dfrac{x+2}{2x}$. Thus the expression is $\dfrac{x+2}{2x(x+2)}=\dfrac1{2x}$ for $x\ne-2$. The limit is $\dfrac1{2(-2)}=-\dfrac14$.

$\dfrac{\sin ax}{bx}=\dfrac{a}{b}\cdot\dfrac{\sin ax}{ax}$. As $x\to0$, $ax\to0$, so the second factor tends to 1. Hence the limit is $\dfrac{a}{b}$.

$\dfrac{\sin ax}{\sin bx}=\dfrac{a}{b}\cdot\dfrac{\sin ax}{ax}\cdot\dfrac{bx}{\sin bx}$. Both standard-limit factors tend to 1, so the limit is $\dfrac{a}{b}$.

Let $u=\pi-x$. Then $u\to0$ and the expression becomes $\dfrac{\sin u}{\pi u}=\dfrac1\pi\cdot\dfrac{\sin u}{u}$. Hence the limit is $\dfrac1\pi$.

Substitute $x=0$: $\dfrac{\cos0}{\pi-0}=\dfrac1\pi$.

Use $\cos2x-1=2\cos^2x-2=2(\cos x-1)(\cos x+1)$. After cancellation, the limit is $\lim_{x\to0}2(\cos x+1)=4$.

Write the expression as $\dfrac{x(a+\cos x)}{b\sin x}=\dfrac{a+\cos x}{b}\cdot\dfrac{x}{\sin x}$. As $x\to0$, this tends to $\dfrac{a+1}{b}$.

Since $\sec x\to1$ as $x\to0$, $x\sec x\to0\cdot1=0$.

As $x\to0$, $\sin ax\sim ax$ and $\sin bx\sim bx$. The numerator is asymptotic to $(a+b)x$ and the denominator is also asymptotic to $(a+b)x$. Since $a+b\ne0$, the limit is $1$.

$\cosec x-\cot x=\dfrac{1}{\sin x}-\dfrac{\cos x}{\sin x}=\dfrac{1-\cos x}{\sin x}$. This equals $\dfrac{2\sin^2(x/2)}{2\sin(x/2)\cos(x/2)}=\tan\dfrac{x}{2}$, whose limit is $0$.

Let $h=x-\dfrac\pi2$. Then $h\to0$ and $2x=\pi+2h$, so $\tan2x=\tan(2h)$. The expression becomes $\dfrac{\tan2h}{h}=2\cdot\dfrac{\tan2h}{2h}\to2$.

At $x=0$, the left-hand expression gives $2(0)+3=3$ and the right-hand expression gives $3(0+1)=3$, so the limit is $3$. At $x=1$, values near 1 use $3(x+1)$, so the limit is $3(2)=6$.

The left-hand limit at 1 is $1^2-1=0$. The right-hand limit is $-1^2-1=-2$. Since the one-sided limits are unequal, $\lim_{x\to1}f(x)$ does not exist.

For $x>0$, $\dfrac{|x|}{x}=1$. For $x<0$, $\dfrac{|x|}{x}=-1$. Since the right-hand and left-hand limits are different, the limit at 0 does not exist.

For $x>0$, $\dfrac{x}{|x|}=1$. For $x<0$, $\dfrac{x}{|x|}=-1$. Since the two one-sided limits are unequal, the limit does not exist.

The absolute value function is continuous at $5$, so the limit is $|5|-5=0$.

For the limit to equal $f(1)=4$, both one-sided limits must be 4. The left-hand limit is $a+b$, and the right-hand limit is $b-a$. Thus $a+b=4$ and $b-a=4$. Solving gives $b=4$ and $a=0$.

If $f(x)=x^2-2$, then $f'(x)=2x$. At $x=10$, $f'(10)=20$.

For $f(x)=x$, $f'(x)=1$ for all $x$. Hence at $x=1$, the derivative is $1$.

For $f(x)=99x$, $f'(x)=99$ for all $x$. Therefore at $x=100$, the derivative is $99$.

Using the first-principle limit gives the same simplified derivatives: (i) $(x^3-27)'=3x^2$. (ii) $(x-1)(x-2)=x^2-3x+2$, so derivative $=2x-3$. (iii) $(x^{-2})'=-2x^{-3}=-\dfrac2{x^3}$. (iv) By simplifying the difference quotient, or equivalently differentiating the quotient, $\left(\dfrac{x+1}{x-1}\right)'=\dfrac{(x-1)-(x+1)}{(x-1)^2}=-\dfrac2{(x-1)^2}$.

Differentiating term by term gives $f'(x)=x^{99}+x^{98}+\cdots+x+1$. Therefore $f'(1)=1+1+\cdots+1$ with 100 terms, so $f'(1)=100$. Also $f'(0)=1$, because all positive powers of $0$ vanish and the constant term in $f'(x)$ is 1. Hence $f'(1)=100=100f'(0)$.

Differentiate term by term, treating $a$ as constant. The derivative of $a^kx^{n-k}$ is $(n-k)a^kx^{n-k-1}$. This gives $nx^{n-1}+(n-1)ax^{n-2}+(n-2)a^2x^{n-3}+\cdots+a^{n-1}$.

(i) $(x-a)(x-b)=x^2-(a+b)x+ab$, so the derivative is $2x-a-b$. (ii) By the chain rule, $\dfrac{d}{dx}(ax^2+b)^2=2(ax^2+b)(2ax)=4ax(ax^2+b)$. (iii) By quotient rule, $\dfrac{(x-b)-(x-a)}{(x-b)^2}=\dfrac{a-b}{(x-b)^2}$.

Using quotient rule with numerator $x^n-a^n$ and denominator $x-a$, the derivative is $\dfrac{nx^{n-1}(x-a)-(x^n-a^n)\cdot1}{(x-a)^2}$. Equivalently, for integer $n$, one may first expand $\dfrac{x^n-a^n}{x-a}=x^{n-1}+ax^{n-2}+\cdots+a^{n-1}$.

Differentiate each expression using power, product and quotient rules. For (ii), $(15x^2+3)(x-1)+(5x^3+3x-1)=20x^3-15x^2+6x-4$. For (vi), $\left(\dfrac{2}{x+1}\right)'=-\dfrac2{(x+1)^2}$ and $\left(\dfrac{x^2}{3x-1}\right)'=\dfrac{2x(3x-1)-3x^2}{(3x-1)^2}=\dfrac{3x^2-2x}{(3x-1)^2}$.

From first principle, $f'(x)=\lim_{h\to0}\dfrac{\cos(x+h)-\cos x}{h}$. Using $\cos(x+h)=\cos x\cos h-\sin x\sin h$, this becomes $\lim_{h\to0}\left[\cos x\dfrac{\cos h-1}{h}-\sin x\dfrac{\sin h}{h}\right]$. Since $\lim_{h\to0}\dfrac{\cos h-1}{h}=0$ and $\lim_{h\to0}\dfrac{\sin h}{h}=1$, the derivative is $-\sin x$.

Use standard trigonometric derivatives and linearity. For (i), $(\sin x\cos x)'=\cos^2x-\sin^2x=\cos2x$. For the remaining parts: $(\sec x)'=\sec x\tan x$, $(\cosec x)'=-\cosec x\cot x$, $(\cot x)'=-\cosec^2x$, $(\sin x)'=\cos x$, $(\cos x)'=-\sin x$, and $(\tan x)'=\sec^2x$.

Using the first-principle limit gives the standard derivatives: $(-x)'=-1$, $((-x)^{-1})'=(-1/x)'=1/x^2$, $(\sin(x+1))'=\cos(x+1)$, and $\left(\cos\left(x-\dfrac\pi8\right)\right)'=-\sin\left(x-\dfrac\pi8\right)$.

Since $a$ is constant, $\dfrac{d}{dx}(x+a)=1$.

Differentiate by product rule: $p\left(\dfrac rx+s\right)+(px+q)\left(-\dfrac r{x^2}\right)=\dfrac{pr}{x}+ps-\dfrac{prx+qr}{x^2}=ps-\dfrac{qr}{x^2}$.

By product and chain rules, derivative $=a(cx+d)^2+(ax+b)\cdot2c(cx+d)=(cx+d)\{a(cx+d)+2c(ax+b)\}$.

Using quotient rule, derivative $=\dfrac{a(cx+d)-c(ax+b)}{(cx+d)^2}=\dfrac{ad-bc}{(cx+d)^2}$.

Simplify $\dfrac{1+1/x}{1-1/x}=\dfrac{x+1}{x-1}$. Differentiating gives $\dfrac{(x-1)-(x+1)}{(x-1)^2}=-\dfrac{2}{(x-1)^2}$.

Write the function as $(ax^2+bx+c)^{-1}$. Its derivative is $-(ax^2+bx+c)^{-2}(2ax+b)$.

Apply quotient rule with numerator derivative $a$ and denominator derivative $2px+q$.

Apply quotient rule with numerator derivative $2px+q$ and denominator derivative $a$.

Write the expression as $ax^{-4}-bx^{-2}+\cos x$. Differentiate term by term to get $-4ax^{-5}+2bx^{-3}-\sin x$.

$4\sqrt{x}-2=4x^{1/2}-2$, so the derivative is $4\cdot\dfrac12x^{-1/2}=\dfrac2{\sqrt{x}}$.

By chain rule, derivative of $(ax+b)^n$ is $n(ax+b)^{n-1}\cdot a=an(ax+b)^{n-1}$.

Use product rule and chain rule on the two factors.

Since $a$ is constant, $\dfrac{d}{dx}\sin(x+a)=\cos(x+a)$.

By product rule, derivative $=(-\cosec x\cot x)\cot x+\cosec x(-\cosec^2x)=-\cosec x\cot^2x-\cosec^3x$.

Using quotient rule, derivative $=\dfrac{(-\sin x)(1+\sin x)-\cos^2x}{(1+\sin x)^2}=\dfrac{-\sin x-1}{(1+\sin x)^2}=-\dfrac1{1+\sin x}$.

Apply quotient rule. The numerator derivative is $\cos x-\sin x$ and the denominator derivative is $\cos x+\sin x$. Simplifying gives $-\dfrac{(\sin x-\cos x)^2+(\sin x+\cos x)^2}{(\sin x-\cos x)^2}=-\dfrac2{(\sin x-\cos x)^2}$.

By quotient rule, derivative $=\dfrac{(\sec x\tan x)(\sec x+1)-(\sec x-1)(\sec x\tan x)}{(\sec x+1)^2}=\dfrac{2\sec x\tan x}{(\sec x+1)^2}$.

Use chain rule on $(\sin x)^n$: derivative $=n(\sin x)^{n-1}\cos x$.

Using quotient rule gives $\dfrac{b\cos x(c+d\cos x)-(a+b\sin x)(-d\sin x)}{(c+d\cos x)^2}$. The numerator simplifies to $bc\cos x+bd(\cos^2x+\sin^2x)+ad\sin x$, i.e. $bc\cos x+ad\sin x+bd$.

Using quotient rule, derivative $=\dfrac{\cos(x+a)\cos x+\sin(x+a)\sin x}{\cos^2x}=\dfrac{\cos((x+a)-x)}{\cos^2x}=\cos a\sec^2x$.

Use product rule. The derivative of $x^4$ is $4x^3$, and the derivative of $5\sin x-3\cos x$ is $5\cos x+3\sin x$.

By product rule, derivative $=(2x)\cos x+(x^2+1)(-\sin x)=2x\cos x-(x^2+1)\sin x$.

Apply product rule. The derivative of $ax^2+\sin x$ is $2ax+\cos x$, and the derivative of $p+q\cos x$ is $-q\sin x$.

Use product rule. The derivative of $x+\cos x$ is $1-\sin x$, and the derivative of $x-\tan x$ is $1-\sec^2x$.

Apply quotient rule with numerator derivative $4+5\cos x$ and denominator derivative $3-7\sin x$.

Since $\cos\dfrac\pi4=\dfrac1{\sqrt2}$, the function is $\dfrac{x^2}{\sqrt2\sin x}$. By quotient rule, derivative $=\dfrac{1}{\sqrt2}\cdot\dfrac{2x\sin x-x^2\cos x}{\sin^2x}$.

Using quotient rule, derivative $=\dfrac{(1)(1+\tan x)-x\sec^2x}{(1+\tan x)^2}$.

Use product rule. The derivative of $x+\sec x$ is $1+\sec x\tan x$, and the derivative of $x-\tan x$ is $1-\sec^2x$.

Write the function as $x\cosec^n x$. Its derivative is $\cosec^n x+x\cdot n\cosec^{n-1}x(-\cosec x\cot x)=\cosec^n x(1-nx\cot x)=\dfrac{1-nx\cot x}{\sin^n x}$.