Limit of $\left(1+\frac{a}{x}\right)^x$ with and without L'Hôpital's rule $$\lim_{x \to \infty} \left(1 +  \frac{a}{x}\right)^{x} = e^{a}$$   
$1$. Without L´Hopital´s rule
$2$. With L´Hopital´s rule
 A: Consider $$\lim_{x\rightarrow \infty}(1+{a\over x})^x.$$ This limit has the indeterminate form $1^\infty$. Let $y=(1+{a\over x})^x$. Taking the natural logarithm of both sides of the equation and simplifying using the rules of logarithms we obtain $\ln(y)=x\ln(1+{a\over x})$. The $$\lim_{x\rightarrow \infty} \ln(y)=\lim_{x\rightarrow \infty}x\ln(1+{a\over x})$$ which has the indeterminate form $\infty\cdot 0$. We can rewrite the right-hand side limit as $$\lim_{x\rightarrow \infty}\ln(y)={\lim_{x\rightarrow \infty}{\ln(1+{a\over x})\over {1\over x}}}$$ which has the indeterminate form ${0\over 0}$. Using L'Hospital's Rule we see that $$\lim_{x\rightarrow \infty}\ln(y)=\lim_{x\rightarrow \infty}{{1\over (1+{a\over x})}\cdot {-a\over x^2}\over {-1\over x^2}}.$$ This simplifies to $$\lim_{x\rightarrow \infty} \ln(y)=\lim_{x\rightarrow \infty}a{1\over (1+{a\over x})}=a.$$ So far we have computed the limit of $\ln(y)$, what we really want is the limit of $y$. We know that $y=e^{\ln(y)}$. So $$\lim_{x\rightarrow \infty}(1+{a\over x})^x=\lim_{x\rightarrow \infty} y=\lim_{x\rightarrow \infty} e^{\ln(y)}=e^a.$$ Thus $$\lim_{x\rightarrow \infty} (1+{a\over x})^x=e^a.$$
A: The function $f(x) = (1+ \frac{a}{x})^x$ is increasing with $x$, so in order to find the limit as $x$ goes to infinity we can assume $x$ is an integer. Then use the binomial theorem to obtain $f(x) = \sum_{k=0}^x \binom{x}{k}(\frac{a}{x})^k$. Let us now estimate when $x$ is very large by looking at the term $\binom{x}{k}(\frac{a}{x})^k = \frac{a^k(x)(x-1)...(x-k+1)}{k!x^k} $ and multiplying out the polynomial in $x$ on top. The first term is $x^k$ and the others don't matter because of the $x^k$ on bottom. Thus (after filling in the details) as $x \rightarrow \infty$ the term becomes $\frac{a^k x^k}{k! x^k} = \frac{a^k}{k!}$. Thus the sum becomes $\sum_{k=0}^\infty \frac{a^k}{k!} = e^a$ as $x$ goes to infinity.
