Differentiation (Part-38) | S N De

 

$\,1.\,$ Give the definition of limit of a function.

Sol. A limit of a function $\,f(x)\,$ can be defined as a value, where the function reaches as the limit reaches some value. The symbol $\,\,\lim_{x \to a}~~f(x)\,\,$ stands for the value of $\,f(x)\,$ when $\,x\,$ is sufficiently close to $\,a.$

$\,2.\,$ Distinction between $\,\lim_{x \to a}\, f(x)\,$ and $\,f(a).$

Sol. The symbol $\,\,\lim_{x \to a}~~f(x)\,\,$ stands for the value of $\,f(x)\,$ when $\,x\,$ is sufficiently close to $\,a.$ In this context, we want to know what happens to the value of $\,f(x)\,$ when $\,x\,$ approaches $\,a\,$ either from the left or from the right of the point $\,x = a.\,$ But we never consider what happens to $\,f(x)\,$ when $\,x\,$ is exactly equal to $\,a\,$. On the contrary $\,f(a)\,$ stands for the value of $\,f(x)\,$ when $\,x\,$ is exactly equal to a. The value of $\,f(a)\,$ is obtained using the definition of $\,f(x)\,$ at $\,x = a\,$ or else by the substitution $\,a\,$ for $\,x\,$ in the analytical expression for $\,f(x)\,$, when it exists.

$\,3.\,$ Using $\,\lim_{x \to 0}\, \frac{e^x-1}{x}=1,\,$ deduce that, $\,\lim_{x \to 0}\,\frac{a^x-1}{x}=\log_e a\,\,[a>0].$

Sol. Let $\,a^x=e^{\log_e a^x}, \\ \log_e a^x=y ~~~(\text{let}) \rightarrow(1) \\ \Rightarrow x \log_e a=y \\ \Rightarrow x=\frac{y}{\log_e a}\rightarrow(2).$

Also, we notice that [from $\,(1)\,$] as $\, x \to 0,~~ y \to 0.$

$\,~~~\lim_{x \to 0}\,\frac{a^x-1}{x}\\=\lim_{x \to 0} \frac{e^{\log_e a^x}-1}{x}\\=\lim_{ y \to 0}\frac{e^y-1}{\frac{y}{\log_e a}}~~[\text{By (2)}]\\=(\log_e a) \times \lim_{y \to 0}\frac{e^y-1}{y}\\=\log_e a \times 1\\=\log_e a\,\,\text{(proved)}$

$\,4.\,$ Starting from $\,~\lim_{x \to a}\frac{x^n-a^n}{x-a}=na^{n-1}\,\,$ deduce that, $\,\lim_{x \to 0}\frac{(1+x)^n-1}{x}=n.$

Sol. We have , $\,~\lim_{x \to a}\frac{x^n-a^n}{x-a}=na^{n-1}\rightarrow(1)$

Let $\,1+x=z\,~~\text{so that as}\,\, x \to 0, z \to 1.$

$\,~~~\lim_{x \to 0}\frac{(1+x)^n-1}{x}\\=\lim_{z \to 1}\frac{z^n-1}{z-1}\\=n \times 1^{n-1}~~[\text{By (1)}]\\=n.$

$\,5.\,$ Using $\,\lim_{x \to 0}\frac{e^x-1}{x}=1,~~$ show that, $~~\lim_{x \to 0}\frac{\log_e(1+x)}{x}=1.$

Sol. We have, $\,\lim_{x \to 0}\frac{e^x-1}{x}=1 \rightarrow(1)$

 Let $\,\log_e(1+x)=z \\ \Rightarrow 1+x=e^z \\ \Rightarrow x=e^z-1.$

Also, we notice that [from $~(1)~$] as $\,x \to 0, z \to 0.$

$~~~~\lim_{x \to 0}\frac{\log_e(1+x)}{x}\\=\lim_{z \to 0} \frac{z}{e^z-1}\\= \frac{1}{\lim_{z \to 0}\left(\frac{e^z-1}{z}\right)}\\=\frac 11~~[\text{Using (1)}]\\=1.$

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