Lim 1 x 1 x e
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We are going to show the following equality: $$ \lim _{x \rightarrow 0}(1+x)^{\frac{1}{x}}=e $$ Firt of all, we definie $u(x)=(1+x)^{\frac{1}{x}}$. We have: $$ \begin{aligned} \ln u(x)&=\ln (1+x)^{\frac{1}{x}}\\ &=\frac{1}{x} \ln (1+x)=\frac{\ln (1+x)}{x}\\ \end{aligned} $$ Two possibilities to find this limit. First: L’Hôpital’s rule. $$ \lim _{x \rightarrow c} \frac{f(x)}{g(x)}=\lim _{x \rightarrow c} \frac{f^{\prime}(x)}{g^{\prime}(x)} $$ Here $c=0$,$f(x)=\ln (1+x)$, $g(x)=x$. Which gives: $$ \lim _{x \rightarrow 0} \frac{ln(1+x)}{x}=\lim _{x \rightarrow 0} \frac{\displaystyle\frac{1}{1+x}}{1}=1 $$ Second: using the definition of the derivative. $$ \begin{aligned} \lim _{x \rightarrow 0} \frac{ln(1+x)}{x}&=\lim _{x \rightarrow 0} \frac{ln(1+x)-ln(1+0)}{x-0} \\ &=\lim _{x \rightarrow 0} \frac{f(x)-f(0)}{x-0}=f^{\prime}(0)=1 \end{aligned} $$ We have: $$ \begin{aligned} \lim _{x \rightarrow 0}(\ln u(x))&=\lim _{x \rightarrow 0} \frac{\ln (1+x)}{x}=1\\ \exp(\lim _{x \rightarrow 0}(\ln u(x))&=\exp (1)\\ \lim _{x \rightarrow 0}(\exp(\ln u(x))&=e \\ \lim _{x \rightarrow 0}(u(x))&=e \\ \end{aligned} $$ We conclude: $$ \lim _{x \rightarrow 0}(1+x)^{\frac{1}{x}}=e $$ Skip to content Home » Blog » Find the limit as x goes to 0 of ((1+x)1/x – e) / x Evalue the limit. First, we have
As
 From this we see that
Hence, using the expansion for
Therefore, we have
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