Question

limT(x-0){(a^x-1)/x then value of limit is

Answer 1

Let $y=a^x-1, a>0,$ then $1+y=a^x,$ we have

$$x=\dfrac{\ln(1+y)}{\ln a}$$ $$\lim\limits_{x\to 0}y=\lim\limits_{x\to 0}(a^x-1)=a^0-1=1-1=0$$

Therefore, the given limit can be written as

$$\lim\limits_{x\to 0}\dfrac{a^x-1}{x}=\lim\limits_{y\to 0}\dfrac{y}{\dfrac{\ln(1+y)}{\ln a}}$$

$$=\ln a\cdot\lim\limits_{y\to 0}\dfrac{1}{\ln(1+y)^{1/y}}=\dfrac{\ln a}{\ln (\lim\limits_{y\to 0}(1+y)^{1/y})}$$

$$=\dfrac{\ln a}{\ln (e)}=\ln a$$

$$\lim\limits_{x\to 0}\dfrac{a^x-1}{x}=\ln a$$