## Exponential

Determine $$\displaystyle \lim_{x\to 0}\frac{e^x-1}{x}$$ with the assumption that $$\displaystyle e=\lim_{n\to \infty}\left(1+\frac1n\right)^n$$

• #### Hint

Express $$e$$ as a limit as $$x$$ approaches 0.

It is better to work with the limit using the inverse function – the natural logarithm.

• #### Resolution

First we should observe that

$$\displaystyle \lim_{x\to 0}\frac{e^x-1}{x} = \left(\lim_{x\to 0}\frac{\ln(x+1)}{x}\right)^{-1}$$.

Letting $$x=\frac 1n$$ we have

$$\displaystyle e= \lim_{n\to \infty}\left(1+\frac1n\right)^n= \lim_{x\to \infty}\left(1+\frac1x\right)^x= \lim_{x\to 0^+}\left(1+x\right)^{\frac1x}$$

Be aware that in this step we replaced the sequence $$\left(1+\frac1n\right)^n$$ with the function $$\left(1+\frac1x\right)^x$$, so we must verify e.g. the continuity and monotonicity of this function on $$(1,\infty)$$ to be sure that the limit of the resulting function exists.

But we need a two-sided limit, so

$$\displaystyle \lim_{x\to 0^-}\left(1+x\right)^{\frac1x}= \lim_{x\to 0^+}\left(1-x\right)^{-\frac1x}= \left(\lim_{x\to 0^+}\left(1-x\right)^{\frac1x}\right)^{-1}= \left(e^{-1}\right)^{-1}=e.$$

Here we used a similar argument as above, except we started with the limit of the sequence $$\lim_{n\to \infty}\left(1-\frac1n\right)^n=e^{-1}$$, which we can of course deduce from $$\lim_{n\to \infty}\left(1+\frac1n\right)^n=e$$.

From here we obtain

$$\displaystyle \lim_{x\to 0}\frac{\ln(x+1)}{x}= \lim_{x\to 0}\ln\left(\exp\left(\frac{\ln(x+1)}{x}\right)\right)= \ln\left(\lim_{x\to 0}\left(1+x\right)^{\frac1x}\right)=\ln e =1$$

Now we must only compute

$$\displaystyle \lim_{x\to 0}\frac{e^x-1}{x} = 1^{-1}=1$$.

and also $$\displaystyle \lim_{x\to 0^-}\frac{e^x-1}{x}= \lim_{x\to 0^-}\frac{\left(\left(1-x\right)^{-\frac1x}\right)^x-1}{x}= \lim_{x\to 0^-}\frac{\frac1{1-x}-1}{x}= \lim_{x\to 0^-}\frac{1-(1-x)}{(1-x)x}=1$$

• #### Result

The limit equals 1.