Limits and Derivatives Test - 9

Given, \(\lim\limits_{x \to \pi} \frac{sin x}{x - \pi} = \lim\limits_{x \to \pi} \frac{sin(\pi - x)}{-(\pi - x)}\)

= -1 \([\because \lim\limits_{x \to 0} \frac{sin x}{x} = 1 \; and \;\pi - x\, as\, x \rightarrow \pi]\)

Given, \(\lim\limits_{x \to 0} \frac{x^2 cos x}{1-cosx}\)

= \(\lim\limits_{x \to 0} \frac{x^2 cos x}{2 sin^2 \frac{x}{2}}[\because 1 - cos x = 2 sin^2 \frac{x}{2}]\)

\(= \lim\limits_{x \to 0} \frac{\frac{x^2}{4} \times 4 cos x}{2sin^2 \frac{x}{2}} = \lim\limits_{x \to 0 } \frac{(\frac{x}{2})^2.2 cosx}{sin^2 \frac{x}{2}}\)

= \(\lim\limits_{\frac{x}{2} \to 0} (\frac{\frac{x}{2}}{sin \frac{x}{2}})^2.2cos x\)

= 2cos 0 = 2 × 1 = 2 \([\because \lim\limits_{x \to 0} \frac{x}{sin x} = 1]\)

Given, \(\lim\limits_{x \to 0} \frac{(1 + x)^n - 1}{x}\) = \(\lim\limits_{x \to 0} \frac{(1 + x)^n - (1)^n}{(1 +x) - (1)}\)

 \(= \lim\limits_{1+x \to 1} \frac{(1 + x)^n - (1)^n}{(1 + x) - (1)} = n(1)^{n-1}\) 

\(= n[\lim\limits_{x \to a} \frac{x^n - a^n}{x-a} = na^{n-1}]\)

Given \(\lim \limits_{x \to 1} \frac{x^m - 1}{x^n - 1}\) = \(\lim\limits_{x \to 1} \frac{\frac{x^m - (1)^m}{x-1}}{\frac{x^n - (1)^n}{x-1}}\)

\(= \frac{m(1)^{m-1}}{n(1)^{n-1}} = \frac{m}{n} [\because \lim\limits_{x \to a} \frac{x^n - a^n}{x-a} = na^{n-1}]\)

Given \(\lim\limits_{\theta \to 0} \frac{1 - cos 4 \theta}{1 - cos 6 \theta} \) 

= \(\lim\limits_{\theta \to 0} \frac{2sin^2 2 \theta}{2 sin^2 3 \theta} [\because 1 - cos \theta = 2sin^2 \frac{\theta}{2}]\)

= \(\lim\limits_{\theta \to 0} \frac{sin^2 2 \theta}{sin^23 \theta} = \lim\limits_{\theta \to 0}[\frac{sin 2 \theta}{sin 3 \theta}]^2\)

= \(\lim\limits_{\theta \to 0 \\ 2\theta \to 0 \\ 3\theta \to 0}[\frac{\frac{sin 2\theta}{2 \theta} \times 2 \theta}{\frac{sin 3 \theta}{3 \theta} \times 3 \theta}]^2\)

= \([\frac{2 \theta}{3\theta}]^2 = (\frac{2}{3})^2 = \frac{4}{9}\)

Given \(\lim\limits_{x \to 0} \frac{cosec x - cot x}{x}\) = \(\lim \limits_{x \to 0} \frac{\frac{1}{sin x} - \frac{cos x}{sin x}}{x}\)

= \(\lim\limits_{x \to 0} \frac{1 -cos x}{x sin x} = \frac{2 sin^2 \frac{x}{2}}{x.2 sin \frac{x}{2} cos \frac{x}{2}}\)

[\(\because\) sin 2x = 2 sin x cos x]

\(= \lim\limits_{x \to 0} \frac{sin \frac{x}{2}}{x cos \frac{x}{2}} = \lim\limits_{x \to 0} \frac{tan \frac{x}{2}}{x}\)

= \( \lim\limits_{x \to 0} \frac{tan \frac{x}{2}}{2 \times \frac{x}{2}}\)

\(= \frac{1}{2} \times 1 = \frac{1}{2} [\because \lim\limits_{x \to 0} \frac{tan x}{x} = 1]\)

Given \( \lim\limits_{x \to 0} \frac{sin x}{\sqrt{x+1} - \sqrt{1-x}}\)

= \(\lim\limits_{x \to 0} \frac{sin x[\sqrt{x +1} + \sqrt{1 - x}]}{(\sqrt{x + 1} - \sqrt{1 - x})(\sqrt{x + 1} + \sqrt{1 -x})}\)

= \(\lim\limits_{x \to 0} \frac{sin x[\sqrt{x +1} + \sqrt{1 - x}]}{x +1 - 1 +x}\)

= \(\lim\limits_{x \to 0} \frac{sin x[\sqrt{x +1} + \sqrt{1 - x}]}{2x}\)

= \(\frac{1}{2}.\lim\limits_{x \to 0} \frac{sin x}{x}[\sqrt{x + 1} + \sqrt {1 - x}]\)

Taking limit, we get

= \(\frac{1}{2} \times 1 \times [\sqrt{0 + 1} + \sqrt{1 - 0}] \)

= \(\frac{1}{2} \times 1 \times 2 = 1\)

Given, \(\lim\limits_{x \to \frac{\pi}{4}} \frac{sec^2 x - 2}{tan x - 1}\) 

= \(\lim\limits_{x \to \frac{\pi}{4}} \frac{1 + tan^2 x - 2}{tan x - 1}\)

= \(\lim\limits_{x \to \frac{\pi}{4}} \frac{tan^2 x - 1}{tan x - 1}\)

= \(\lim\limits_{x \to \frac{\pi}{4}} \frac{(tan x + 1)(tan x - 1)}{(tan x - 1)}\)

= \(\lim\limits_{x \to \frac{\pi}{4}}\) (tan x + 1) = tan \(\frac{\pi}{4}\) + 1

= 1 + 1 = 2

Given, \(\lim\limits_{x \to 1} \frac{(\sqrt x - 1)(2x -3)}{2x^2 + 3x -2x -3}\)

= \(\lim\limits_{x \to 1} \frac{(\sqrt x - 1)(2x -3)}{x(2x + 3) - 1(2x +3)}\)

= \(\lim\limits_{x \to 1} \frac{(\sqrt x - 1)(2x -3)}{(x -1)(2x + 3)}\)

= \(\lim\limits_{x \to 1} \frac{(\sqrt x - 1)(\sqrt x + 1)(2x -3)}{(x-1)(\sqrt x + 1)(2x+3)}\)

= \(\lim\limits_{x \to 1} \frac{(x - 1)(2x -3)}{(x-1)(\sqrt x + 1)(2x+3)}\)

= \(\lim\limits_{x \to 1} \frac{2x -3}{(\sqrt x + 1)(2x+3)}\)

Taking limit, we have

\(= \frac{2(1) -3}{(\sqrt 1 + 1)(2 \times 1 + 3)} = \frac{-1}{2 \times 5} = \frac{-1}{10}\)

Given, f(x) = \( \begin{cases} \frac{sin[x]}{[x]}, & \quad [x] \neq 0 \\ 0, & \quad [x] = 0 \end{cases} \)

LHL = \(\lim\limits_{x \to 0^-} \frac{sin[x]}{[x]} = \lim\limits_{h \to 0} \frac{sin[0 - h]}{[0 - h]} \)

\(= \lim\limits_{h \to 0} \frac{sin(-1)}{-1} = \frac{-sin(1)}{-1}=sin(1)\)

\(\because \lim\limits_{x \to 0^-} {[x]}=\lim\limits_{h \to 0} {[0+h]}=\lim\limits_{h \to 0}0=0\)

\(\therefore \) \(f(x)=0 \, when \, limit \,x \,tends\, to \,0\)

\(\therefore \)RHL = \(\lim\limits_{x \to 0^+} f(x)=\lim\limits_{x \to 0^+}0=0.\)

LHL ≠ RHL

So, the limit does not exist.

Given, \(\lim\limits_{x \to 0} \frac{|sin x|}{x} \) 

LHL = \(\lim\limits_{x \to 0} \frac{-sin x}{x} = -1 [\therefore \lim\limits_{x \to 0} \frac{sin x}{x} = 1]\)

RHL = \(\lim\limits_{x \to 0} \frac{sin x}{x} = 1\)

LHL ≠ RHL, so the limit does not exist.

Given \(f(x) = \begin{cases} x^2 - 1, & \quad 0 < x < 2 \\ 2x + 3, & \quad 2 \leq x < 3 \end{cases}\)

\(\therefore \lim\limits_{x \to 2^-} f(x) = \lim\limits_{x \to 2^-}(x^2 - 1)\)

= \(\lim\limits_{h \to 0}[(2 - h)^2 - 1] = \lim\limits_{h \to 0}(4 + h^2 - 4h - 1)\)

= \(\lim\limits_{h \to 0}(h^2 - 4h + 3) = 3\)

and \(\lim\limits_{x \to 2^+}f(x) = \lim\limits_{x \to 2^+}(2x + 3)\)

= \(\lim\limits_{h \to 0}\)[2(2 + h) + 3] = 7

Therefore, the quadratic equation whose roots are 3 and 7 is \(x ^2\) – (3 + 7)x + 3×7 = 0 i.e., \(x ^2\) – 10x + 21 = 0.

Given, \(\lim\limits_{x \to 0} \frac{tan 2x - x}{3x - sin x}\) 

= \(\lim\limits_{x \to 0} \frac{x[\frac{tan 2x}{x} - 1]}{x[3 - \frac{sin x}{x}]}\)

= \(\lim\limits_{x \to 0 \\ \therefore 2x \to 0} \frac{\frac{tan2x}{2x} \times 2 - 1}{3 - \frac{sin x}{x}} = \frac{1.2 - 1}{3 - 1}\)

= \(\frac{2-1}{2} = \frac{1}{2}\)

Given f(x) = x – [x]

we have to first check for differentiability of f(x) at x = \(\frac{1}{2}\)

\(\therefore Lf' (\frac{1}{2})\) = LHD = \(\lim\limits_{h \to 0} \frac{f[\frac{1}{2} - h] - f [\frac{1}{2}]}{-h}\)

= \(\lim\limits_{h \to 0} \frac{(\frac{1}{2} -h) - [\frac{1}{2} - h] - \frac{1}{2} + [\frac{1}{2}]}{-h}\)

= \(\lim\limits_{h \to 0} \frac{\frac{1}{2} - h - 0 - \frac{1}{2} + 0}{-h} = \frac{-h}{-h} = 1\)

Now, Rf'\((\frac{1}{2}) = RHD = \lim\limits_{h \to 0} \frac{f(\frac{1}{2} + h) - f(\frac{1}{2})}{h}\)

= \(\lim\limits_{h \to 0} \frac{(\frac{1}{2} + h) - [\frac{1}{2} + h] - \frac{1}{2} + [\frac{1}{2}]}{h}\)

= \(\lim\limits_{h \to 0} \frac{\frac{1}{2} +h - 0 - \frac{1}{2} + 0}{h} = \frac{h}{h} = 1\)

Since LHD = RHD

\(\therefore f'(\frac{1}{2}) = 1\)

Given that \(y = \sqrt x + \frac{1}{\sqrt x}\)

\(\frac{dy}{dx} = \frac{1}{2\sqrt x} - \frac{1}{2x^{\frac{3}{2}}}\)

\((\frac{dy}{dx})_{at\, x = 1} = \frac{1}{2} - \frac{1}{2} = 0\)

Given that f(x) = \(\frac{x - 4}{2 \sqrt x}\)

\(\therefore\) f'(x) = \(\frac{1}{2}[\frac{\sqrt x.1 - (x - 4) . \frac{1}{2\sqrt x}}{x}]\)

= \(\frac{1}{2}[\frac{2x - x + 4}{2\sqrt x.x}] = \frac{1}{2}[\frac{x+4}{2(x)^{\frac{3}{2}}}]\)

∴ f’(x) at x = 1 = \(\frac{1}{2}[\frac{1+4}{2\times1}] = \frac{5}{4}\)

Given, \(y = \frac{1 + \frac{1}{x^2}}{1 - \frac{1}{x^2}}\) ⇒ \(y = \frac{x^2 + 1}{x^2 - 1}\)

\(\therefore \frac{dy}{dx} = \frac{(x^2 - 1).2x - (x^2 + 1).2x}{(x^2 - 1)^2}\)

\(= \frac{2x(x^2 - 1 - x^2 - 1)}{(x^2 - 1)^2} = \frac{2x(-2)}{(x^2 - 1)^2}\)

= \(\frac{-4x}{(x^2 - 1)^2}\)

Given \(y = \frac{sin x + cosx}{sinx - cosx}\)

\(\frac{dy}{dx} = \frac{(sin x - cos x)(cos x - sin x) - \\ (sin x + cos x)(cos x + sin x)}{(sin x - cos x)^2}\)

\(= \frac{-(sin x - cos x)^2 - (sin x + cos x)^2}{(sin x - cos x)^2}\)

 \(= \frac{-[sin^2 x + cos^2x - 2sin xcosx \\ + sin^2x + cos^2x + 2sinx cos x]}{(sin x - cosx)^2}\)

\(= \frac{-2}{(sin x - cos x)^2}\)

\(\therefore (\frac{dy}{dx})_{at\, x = 0} = \frac{-2}{(sin 0 - cos 0)^2} \)

= \(\frac{-2}{(-1)^2} = -2\)

Given \(y = \frac{sin(x + 9)}{cos x}\)

\(\frac{dy}{dx} = \frac{cosx.cos(x + 9) - sin(x + 9)(-sin x)}{cos^2x}\)

\(= \frac{cosxcos(x + 9)+sinxsin(x+9)}{cos^2x}\)

= \(\frac{cos(x + 9 - x)}{cos^2x} = \frac{cos 9}{cos^2 x}\)

\(\therefore (\frac{dy}{dx})_{at\, x\, =\, 0} = \frac{cos9}{cos^20} = \frac{cos9}{(1)^2}\)

= cos 9

Given f(x) = \(1 + x + \frac{x^2}{2} + \dots + \frac{x^{100}}{100}\)

f’(x) = \(1 + \frac{2x}{2} + \dots + \frac{100x^{99}}{100}\)

∴ f’(1) = 1 + 1 + 1 +…+ 1 (100 times) = 100