How do I prove that this limit is equal to e without L'Hospital?

Question

I'm solving some limits and in one of my examples I need to use the fact that:

I am, however, unable to prove that this is actually true. I believe I can't just substitue t=(1/x) because than we can not say where such t is going. I tried to find the proof online, but they all rely on L'Hôpital's rule, which I am trying to avoid. How can I prove this equality without it?

Answer 1

We will first prove a supporting result using sandwich theorem

$$ \begin{align} \frac{x}{x+1} &< \ln(1+x) < x &\text{for $x > -1$} \end{align} $$ See this reference.

$$ \lim_{x \to 0} \frac{1}{x+1} < \lim_{x \to 0} \frac {\ln(1+x)}{x} < \lim_{x \to 0} 1 $$

$$ \Rightarrow \lim_{x \to 0} \frac {\ln(1+x)}{x}=1 $$

And now we can solve the limit in the question. $$ \lim_{x \to 0}(1+x)^{\frac{1}{x}} = e^{\lim_{x \to 0}\frac {\ln(1+x)}{x}}=e $$

Answer 2

Write it as $(1+\frac{1}{n})^n$ and get the binomial expansion $\sum_{k=0}^n\frac{n(n-1)...(n-k+1)}{n^k}\frac{1}{k!}$. As $n\to \infty$, expansion converges to infinite series for $e=\sum_{k=0}^\infty\frac{1}{k!}$

Note: this is way I first learnt it in high school.

Answer 3

Let $g(y) = \frac 1y$. $\lim_{y\to \infty} g(y) = 0$. Thus if $\lim_{k\to\infty} f(k) = \lim_{g(k)\to 0} f(k)$.

$\lim_{x\to 0}(1 + x)^{\frac 1x}$. Replace $x$ with $g(y)$

$\lim_{g(y)\to 0}(1+g(y))^{\frac 1{g(y)}}=$

$\lim_{g(y)\to 0}(1+\frac 1y)^y=$

$\lim_{y\to \infty}(1+\frac 1y)^y =e$