Differentiation (Part-38) | S N De

 

$\,12.\,$ Show :

$\,(i)\,\lim_{x  \to 2}\,\frac{x^9-512}{x^4-16}=72$

Sol. $\,\,\lim_{x  \to 2}\,\frac{x^9-512}{x^4-16}\\=\lim_{x  \to 2}\frac{x^9-2^9}{x^4-2^4}\\=\frac{\lim_{x  \to 2}\frac{x^9-2^9}{x-2}}{\lim_{x  \to 2}\frac{x^4-2^4}{x-2}}~~[*]\\=\frac{9 \times 2^{9-1}}{ 4\times 2^{4-1}}\\=\frac 94 \times \frac{2^8}{2^3}\\=\frac{9}{2^2} \times 2^5\\=9 \times 2^3\\=72\,\,\text{(proved)}$

Note[*]: $\,\lim_{x \to a}~~\left(\frac{x^n-a^n}{x-a}\right)=na^{n-1}$

$\,(ii)\,\lim_{x  \to 2}\frac{(1+x)^n-3^n}{x-2}=n \cdot 3^{n-1}$

Sol. $\,\lim_{x  \to 2}\,\frac{(1+x)^n-3^n}{x-2}\\=\lim_{x  \to 2}\frac{(1+x)^n-3^n}{(1+x)-3}\\=\lim_{z  \to  3}\frac{z^n-3^n}{z-3}\\~~~[\text{let}\,\,z=1+x,~~\text{so as}~~x \to 2,\,z \to 3.]\\=n\cdot 3^{n-1}\,\,\text{(proved)}$

$\,(iii)\,\lim_{x  \to  3}\frac{x^{-6}-3^{-6}}{x^{-4}-3^{-4}}=\frac 16$

Sol. $\,~~\lim_{x  \to  3}\frac{x^{-6}-3^{-6}}{x^{-4}-3^{-4}}\\=\lim_{x  \to  3}\frac{\frac{x^{-6}-3^{-6}}{x-3}}{\frac{x^{-4}-3^{-4}}{x-3}}\\=\frac{-6 \times 3^{-6-1}}{-4 \times 3^{-4-1}}\\=\frac 64 \times \frac{3^{-7}}{3^{-5}}\\=\frac 32 \times 3^{-7+5}\\=\frac 32 \times 3^{-2}\\=\frac{1}{2} \times 3^{-1}\\=\frac 16\,\,\text{(proved)}$

$\,(iv)\,\lim_{h \to 0}\,\frac{\sqrt[3]{h+1}-1}{h}=\frac 13$

Sol. $\,~~~\lim_{h \to 0}\,\frac{\sqrt[3]{h+1}-1}{h}\\=\lim_{h \to 0}\frac{(\sqrt[3]{h+1}-1)\{(h+1)^{\frac 23}+(h+1)^{\frac 13}+1\}}{h\{(h+1)^{\frac 23}+(h+1)^{\frac 13}+1\}}\\=\lim_{h \to 0}\frac{(\sqrt[3]{h+1})^3-1^3}{h\{(h+1)^{\frac 23}+(h+1)^{\frac 13}+1\}}\\=\lim_{h \to 0}\frac{h+1-1}{h\{(h+1)^{\frac 23}+(h+1)^{\frac 13}+1\}}\\=\lim_{h \to 0}\frac{h}{h\{(h+1)^{\frac 23}+(h+1)^{\frac 13}+1\}}\\=\lim_{h \to 0}\frac{1}{\{(h+1)^{\frac 23}+(h+1)^{\frac 13}+1\}}\\=\frac{1}{\{(0+1)^{\frac 23}+(0+1)^{\frac 13}+1\}}\\=\frac{1}{1+1+1}\\=\frac 13\,\,\text{(proved)}$

$\,(v)\lim_{x  \to  a}\,\frac{\sqrt[3]x-\sqrt[3]a}{x-a}=\frac 13 a^{-2/3}$

Sol. $\lim_{x  \to  a}\frac{\sqrt[3] x-\sqrt[3] a}{x-a}\\=\lim_{x  \to  a}\frac{(\sqrt[3] x-\sqrt[3] a)(x^{2/3}+(xa)^{1/3}+a^{2/3})}{(x-a)(x^{2/3}+(xa)^{1/3}+a^{2/3})}\\=\lim_{x  \to  a}\frac{(\sqrt[3]x)^3-(\sqrt[3]a)^3}{(x-a)(x^{2/3}+(xa)^{1/3}+a^{2/3})}\\=\lim_{x  \to  a}\frac{(x-a)}{(x-a)(x^{2/3}+(xa)^{1/3}+a^{2/3})}\\=\lim_{x  \to  a}\frac{1}{(x^{2/3}+(xa)^{1/3}+a^{2/3})}\\=\frac{1}{a^{2/3}+(a \times a)^{1/3}+a^{2/3}}\\=\frac{1}{a^{2/3}+a^{2/3}+a^{2/3}}\\=\frac{1}{3a^{2/3}}\\=\frac 13 a^{-2/3}\,\,\text{(proved)}$

$\,(vi)\lim_{x \to a}\,\frac{x^{\frac 38}-a^{\frac 38}}{x^{\frac 53}-a^{\frac 53}}$

Sol. $\,\lim_{x \to a}\frac{x^{\frac 38}-a^{\frac 38}}{x^{\frac 53}-a^{\frac 53}}\\=\frac{\lim_{x \to a}\frac{x^{\frac 38}-a^{\frac 38}}{x-a}}{\lim_{x \to a}\frac{x^{\frac 53}-a^{\frac 53}}{x-a}}\\=\frac{\frac 38 \times a^{\frac 38-1}}{\frac 53\times a^{\frac 53-1}}\\=\frac 38 \times \frac 35 \times \frac{a^{-5/8}}{a^{2/3}}\\=\frac{9}{40}a^{-\frac 58-\frac 23}\\=\frac{9}{40}a^{-\frac{31}{24}}~~\text{(ans.)}$

$\,(vii)\,\lim_{x \to 0} \frac 1x[(1+x)^8-1]=8$

Sol. $\,\lim_{x \to 0}\frac 1x[(1+x)^8-1]\\=\lim_{x \to 0}\frac{(1+x)^8-1}{(1+x)-1}\\=\lim_{z \to 1} \left(\frac{z^8-1}{z-1}\right)\\~~~[\text{let}\,\,z=x+1,\,\,\text{so that as}~~ x \to 0, z \to 1.]\\=8\times 1^{8-1}\\=8\,\,\text{(proved)}$

$\,(viii)\lim_{x \to 0}\,\frac{(1+x)^9-1}{(1+x)^6-1}=\frac 32$

Sol. $\,\lim_{x \to 0}\frac{(1+x)^9-1}{(1+x)^6-1}\\=\frac{\lim_{x \to 0}\frac{(1+x)^9-1}{x}}{\lim_{x \to 0}\frac{(1+x)^6-1}{x}}\\=\frac 96~~[\because \lim_{x \to 0} \frac{(1+x)^n-1}{x}=n]\\=\frac 32\,\,\text{(proved)}$

$\,13.\,$ If $\,f(x)=ax^2+bx+c,\,$ show that , $\,\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}=2ax+b.$

Sol. $\,f(x+h)-f(x)\\=[a(x+h)^2+b(x+h)+c]\\~~~~~~~~~-(ax^2+bx+c)\\=[a(x^2+2xh+h^2)+bx+bh+c]\\~~~~~~~~~~~-ax^2-bx-c\\=ax^2+2axh+ah^2+bx+bh+c\\~~~~~~~~~~~~~~-ax^2-bx-c\\=h(2ax+b) \\ \Rightarrow \frac{f(x+h)-f(x)}{h}=2ax+b \\ \Rightarrow \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}=2ax+b\,\,\text{(proved)}$

$\,14.\,$ If $\,f(x)=\frac 1x,\,\,$ prove that, $\,\lim_{h \to 0}\frac{f(2+h)-f(2)}{h}=-\frac 14.$

Sol. $\,f(2+h)-f(2)\\=\frac{1}{2+h}-\frac 12\\=\frac{2-2-h}{2(2+h)}\\=\frac{-h}{2(2+h)}\\ \Rightarrow \frac{f(2+h)-f(2)}{h}=\frac{-1}{2(2+h)}\\\Rightarrow \lim_{h \to 0}\frac{f(2+h)-f(2)}{h}\\=\lim_{h \to 0}\frac{-1}{2(2+h)}\\=\frac{-1}{2(2+0)}\\=-\frac 14\,\,\text{(proved)}$