Differentiation (Part-38) | S N De

 

$\,8.\,$ Prove :

$\,(i)\,~~\lim_{x \to 0}~~ x\cos \frac 1x=0$

Sol.  We know that  the cosine function is bounded by $\,-1\,$ and $\,1\,$:

$\, -1<\cos \frac 1x <1.$

And we know that  $~~~\lim_{x \to 0} x=0.$

Therefore, the limit of a bounded function mulitplied by a function that goes to $\,0\,$ must be $\,0.$

$\,\therefore~~\lim_{x \to 0}~~ x\cos \frac 1x=0.$

$\,8.\,$ Prove :

$\,(ii)\,~~\lim_{x \to 0}~~ x^2\sin \frac 1x=0$

Sol.  We know that  the sine function is bounded by $\,-1\,$ and $\,1\,$:

$\, -1<\sin \frac 1x <1.$

And we know that  $~~~\lim_{x \to 0} x^2=0.$

Therefore, the limit of a bounded function mulitplied by a function that goes to $\,0\,$ must be $\,0.$

$\,\therefore~~\lim_{x \to 0}~~ x^2\sin \frac 1x=0.$

$\,(iii)\,\lim_{x \to 0}~~\frac{x^2\sin\frac 1x}{\sin x}=0$

Sol. $\,\, \lim_{x \to 0}~~\frac{x^2\sin\frac 1x}{\sin x}\\=\lim_{x \to 0}\frac{x \sin\frac 1x}{\frac{\sin x}{x}}\\=\frac 01\\=0\,\,\text{(proved)}$

$\,(iv)\,\,\lim_{x \to \infty}\frac{\sin x}{x}=0$

Sol. Let $\,\,L=\lim_{x \to \infty}\frac{\sin x}{x}\\~~~~=\lim_{z \to 0} z\sin\frac 1z~~[\text{where}~~x=\frac 1z]\rightarrow(1)$

Now, we'll prove that $\,\,~~\lim_{z \to 0}~~ z\sin \frac 1x=0$

 We know that  the sine function is bounded by $\,-1\,$ and $\,1\,$:

$\, -1<\sin \frac 1z <1.$

And we know that  $~~~\lim_{z \to 0} z=0.$

Therefore, the limit of a bounded function mulitplied by a function that goes to $\,0\,$ must be $\,0.$

$\,\therefore~~\lim_{z \to 0}~~ z\sin \frac 1z=0.$

Hence, by $\,(1),\,$ we get, $~~L=0.$

$\,(v)\, \lim_{x \to \infty}~~\frac{\cos x}{x}=0$

Sol. Let $\,\,L=\lim_{x \to \infty}~~\frac{\cos x}{x}\\~~~~=\lim_{z \to 0}z \cos \frac 1z\,\,[\text{where}\,\,z=\frac 1x]\rightarrow(1)$

Now, using the result in $\,8(i),\,$ we can conclude from $\,(1),\, L=0.$

$\,9.\,$ Putting $\,z=x-\frac{\pi}{4},\,\,$ show that $\lim_{x \to \pi/4}\frac{1-\tan x}{1-\sqrt 2 \sin x}=2.$

Sol.  $\lim_{x \to \pi/4}\frac{1-\tan x}{1-\sqrt 2 \sin x}\\=\lim_{z \to 0}\frac{1-\tan(z+\pi/4)}{1-\sqrt2 \sin(z+\pi/4)}\\=\lim_{z \to 0}\frac{\frac{-2\tan z}{1-\tan z}}{1-\cos z-\sin z }~~[*]\\=\lim_{z \to 0}\frac{-2\tan z}{(1-\tan z)\left(2\sin^2\frac z2-2\sin\frac z2\cos \frac z2\right)}\\=\lim_{z \to 0} \frac{-2\frac{\sin z}{\cos z}}{(1-\tan z)\cdot 2\sin \frac z2\left(\sin\frac z2-\cos\frac z2\right)}\\=\lim_{z \to 0}\frac{-\sin z}{\cos z(1-\tan z)\sin \frac z2\left(\sin\frac z2-\cos\frac z2\right)}\\=\lim_{z \to 0}\frac{-2\sin \frac z2 \cos\frac z2}{\cos z (1-\tan z)\sin \frac z2\left(\sin\frac z2-\cos\frac z2\right)}\\=\lim_{z \to 0}\frac{-2\cos \frac z2}{\cos z(1-\tan z)\left(\sin\frac z2-\cos\frac z2\right)}\\=\frac{-2 \times 1}{1(1-0)(0-1)}\\=2\,\,\text{(showed)}$

Note[*] : $~~~~1-\tan(z+\pi/4)\\=1-\frac{\tan z+\tan \pi/4}{1-\tan z\tan \pi/4}\\=1-\frac{\tan z+1}{1-\tan z}\\=\frac{1-\tan z -\tan z-1}{1-\tan z}\\=\frac{-2\tan z}{1-\tan z}\\ 1-\sqrt 2\sin(z+\pi/4)\\=1-\sqrt 2(\sin z\cos \pi/4+\cos z \sin \pi/4)\\=1-\sqrt 2 \times \frac{1}{\sqrt 2}(\sin z+\cos z)\\=1-\sin z-\cos z.$

$\,10.\,$ Putting $\,x=\theta-\pi/4,\,\,$ prove that $\,\lim_{\theta \to \pi/4}\frac{\sin \theta-\cos\theta}{\theta-\pi/4}=\sqrt 2.$

Sol. $\,\lim_{\theta \to \pi/4}\frac{\sin \theta-\cos\theta}{\theta-\pi/4}\\=\lim_{\theta \to \pi/4}~~\sqrt 2. \left(\frac{\frac{1}{\sqrt 2}\sin\theta-\frac{1}{\sqrt 2}\cos \theta}{\theta-\pi/4}\right)\\=\sqrt2.\lim_{\theta \to \pi/4}\left[\frac{\sin(\theta-\pi/4)}{\theta-\pi/4}\right]\\=\sqrt{2}.\lim_{x \to 0} \frac{\sin x}{x} \\~~~[\text{as}\,\, \theta \to \pi/4,\,\, x \to 0\,~~\text{since}~~x=\theta-\pi/4]\\=\sqrt 2 \times 1\\=\sqrt 2\,\,\text{(proved)}$

$\,11.\,$ If $\,\,\lim_{x \to 2}\frac{ax^2-b}{x-2}=4,\,\,$ find the values of $\,a\,$ and $\,b.$

Sol. Let $\,\,(ax^2-b)=(x-2)(cx+d) \rightarrow(1)$

Now, comparing both sides of $\,(1),\,$ we get,

$\,a=c,\,d=2c,\,b=2d \rightarrow (2)$

Hence, $\,\,\lim_{x \to 2}\frac{ax^2-b}{x-2}=4 \\ \Rightarrow \lim_{x \to 2}\frac{(x-2)(cx+d)}{x-2}=4\\ \Rightarrow 2c+d=4 \\ \therefore c=1,\,\,d=2 \rightarrow (3)$

So, by $\,(2)\,$ and $\,(3)\,$, we get,

$\,a=1,\,b=2\times 2=4 \,\,\text{(ans.)}$

$\,12.\,$ Prove that :

$\,(i)\,\lim_{x \to 1}\frac{4^x-4}{x-1}=8 \log_e 2$

$\,\, \lim_{x \to 1}\frac{4^x-4}{x-1}\\=\lim_{x \to 1}\frac{4(4^{x-1}-1)}{x-1}\\=4 \times \lim_{z \to 0}\frac{4^z-1}{z}~~~[*]\\ =4 \times \log_e 4\\=4 \times \log_e 2^2\\ =4 \times 2\log_e 2\\=8\log_e 2\,\,\text{(proved)}$

Note[*] : $[\text{where}~~z=x-1,\,\,\text{so}~~z \to 0\,\,\text{as}~~x \to 1]$

$\,12(ii).\, \lim_{x \to 2} \frac{\log(2x-3)}{2(x-2)}=1$

Sol. $\,\,\,\lim_{x \to 2} \frac{\log(2x-3)}{2(x-2)}\\=\lim_{z \to 0} \frac{\log[2(z+2)-3]}{2z}~~[*]\\=\lim_{z \to 0} \frac{\log(2z+1)}{2z}\\=\lim_{p \to 0} \frac{\log(p+1)}{p}~~~[p=2z]\\=1\,\,\text{(proved)}$

Note[*] : $\,z=x-2,\,\,\text{and so, as}~~x \to 2, z \to 0.$