Finding limit without using limit

If we have to find the value of $$ \lim_{x \to 0} \frac{e^x-1}{x}$$

I tried to solve this by using series i.e by expanding $e^x$ and got the result.

But if there is another method to solve this

If one knows that $$ \left( e^x\right)'=e^x,\quad x \in \mathbb{R}, $$ then, using the definition, for $f$ derivable near $a$, $$ \lim_{x \to a}\frac{f(x)-f(a)}{x-a}=f'(a) $$ yields

$$ \lim_{x \to a}\frac{e^x-1}{x}=\lim_{x \to a}\frac{e^x-e^0}{x-0}=e^0=1 $$ as wanted.

There is a simpler way to calculate the limit provided we know the following two facts:

• $e^{x} \geq 1 + x$ for all $x$.
• $e^{x + y} = e^{x}\cdot e^{y}$ for all $x, y$.

For the derivation here we need the first inequality $e^{x} \geq 1 + x$ to hold for all $x \in (-1, \infty)$ and not on whole of $\mathbb{R}$. Also we need the second property only for the specific case when $y = -x$ so that $e^{x}e^{-x} = 1$.

Both these facts can be established with some effort (see second half of the answer) via definition $$e^{x} = \lim_{n \to \infty}\left(1 + \frac{x}{n}\right)^{n}\tag{1}$$ I now prove that $(e^{x} - 1)/x \to 1$ as $x \to 0$. First we handle the case when $x \to 0^{+}$. The inequality $e^{x} \geq 1 + x$ shows that $$\frac{e^{x} - 1}{x}\geq 1\tag{2}$$ Further let $0 < x < 1$ and then we have the inequality $$e^{-x} \geq 1 - x$$ We can take reciprocals to get $$e^{x} \leq \frac{1}{1 - x}$$ Here we make use of the fact that $e^{x + y} = e^{x}e^{y}$. And this inequality gives us $$\frac{e^{x} - 1}{x} \leq \frac{1}{1 - x}\tag{3}$$ From equations $(2)$ and $(3)$ we get $$1 \leq \frac{e^{x} - 1}{x} \leq \frac{1}{1 - x}$$ for $0 < x < 1$ and letting $x \to 0^{+}$ and using Squeeze theorem we get $$\lim_{x \to 0^{+}}\frac{e^{x} - 1}{x} = 1\tag{4}$$ Note that the above limit implies that $e^{x} \to 1$ as $x \to 0^{+}$.

Next we handle the case when $x \to 0^{-}$. We put $x = -y$ to get $$\lim_{x \to 0^{-}}\frac{e^{x} - 1}{x} = \lim_{y \to 0^{+}}\frac{e^{-y} - 1}{-y} = \lim_{y \to 0^{+}}\frac{e^{y} - 1}{y}\cdot\frac{1}{e^{y}} = 1\cdot 1 = 1\tag{5}$$ and using equations $(4)$ and $(5)$ we have $$\lim_{x \to 0}\frac{e^{x} - 1}{x} = 1$$

For those who are interested in the proof of the properties of $e^{x}$ mentioned at the start of my answer, I provide an outline (with some details to be filled by the reader). Before we can proceed to establish the properties of $e^{x}$ it is necessary to show that the definition of $e^{x}$ via limit in $(1)$ makes sense.

It is easy to prove (and this is available in many textbooks and also on MSE) that the following limit $$\lim_{n \to \infty}\left(1 + \frac{1}{n}\right)^{n}\tag{6}$$ exists and is normally denoted by $e$. Moreover $2 < e < 3$. Using the existence of limit $(6)$ and a little amount of algebraic manipulation it is easy to prove that if $x$ is a positive integer then $$\lim_{n \to \infty}\left(1 + \frac{x}{n}\right)^{n} = e^{x}$$ For a real number $x$ we need some more work. For $x > 0$ it is easy to see via binomial theorem that sequence $(1 + (x/n))^{n}$ is increasing and further if $r$ is a positive integer with $r > x$ then $(1 + (x/n))^{n} < (1 + (r/n))^{n}$. Now $(1 + (r/n))^{n}$ is also increasing and tends to $e^{r}$ hence it follows that $(1 + (r/n))^{n} \leq e^{r}$ for all positive integers $n$. Hence the sequence $(1 + (x/n))^{n}$ is also bounded above by $e^{r}$ and thus tends to a limit. It follows that the limit $$F(x) = \lim_{n \to \infty}\left(1 + \frac{x}{n}\right)^{n}$$ exists for all $x \geq 0$ and moreover since the sequence is increasing the limit is not less than the value of sequence at $n = 1$. Thus we see that $$F(x) = \lim_{n \to \infty}\left(1 + \frac{x}{n}\right)^{n} \geq 1 + x\tag{7}$$ for all $x \geq 0$. Thus $F(x) > 0$ for all $x \geq 0$.

For $x < 0$ let us put $y = -x$ so that $y > 0$. The limit $$F(y) = \lim_{n \to \infty}\left(1 + \frac{y}{n}\right)^{n}\tag{8a}$$ exists and is positive. Further we know via Bernoulli's Inequality that if $n > |x|$ then $$1 - \frac{x^{2}}{n} \leq \left(1 - \frac{x^{2}}{n^{2}}\right)^{n} \leq 1$$ and hence on taking limit when $n \to \infty$ we get via Squeeze Theorem $$\lim_{n \to \infty}\left(1 - \frac{x^{2}}{n^{2}}\right)^{n} = 1\tag{8b}$$ Dividing $(8b)$ by $(8a)$ (and noting that $y = -x$) we see that the limit $$\lim_{n \to \infty}\left(1 + \frac{x}{n}\right)^{n}$$ exists and is positive. Thus we have proved that the limit $$F(x) = \lim_{n \to \infty}\left(1 + \frac{x}{n}\right)^{n}$$ exists for all $x$ and is positive and the definition $(1)$ is now justified. Moreover equation $(8b)$ shows that $F(x)F(-x) = 1$ for all $x$. Therefore we have $$F(-x) = 1/F(x) = \lim_{x \to \infty}\left(1 + \frac{x}{n}\right)^{-n}$$ and using binomial theorem for general index and assuming $0 < x < 1$ we can prove that $F(-x) \geq 1 - x$. Combined with equation $(7)$ it follows that $F(x) \geq 1 + x$ for all $x \in (-1, \infty)$ and this establishes the first property of $F(x)$. Also note that we have proved the specific case of second property namely $F(-x)F(x) = 1$ which was required in our solution, but it is easy to prove in general that $F(x + y) = F(x)F(y)$. For this we use the following lemma:

Lemma 1: If $x_{n}$ is a sequence of real or complex terms such that $n(x_{n} - 1)\to 0$ as $n \to \infty$ then $x_{n}^{n} \to 1$ as $n \to \infty$.

This lemma is proved in this answer. Now let us define the sequence $x_{n}$ by $$x_{n} = \dfrac{\left(1 + \dfrac{x + y}{n}\right)}{\left(1 + \dfrac{x}{n}\right)\left(1 + \dfrac{y}{n}\right)}$$ and it is easy to check that $n(x_{n} - 1) \to 0$ and therefore $x_{n}^{n} \to 1$ so that $F(x + y) = F(x)F(y)$ for all $x, y$.

The sequence $\left(1+\frac{x}{n}\right)^n$ increases to $e^x$ thus setting $n=1$ we get

$$1+x\leq e^x$$ which gives

$$1\leq \frac{e^x-1}{x}$$

Similarly the sequence $\left(1+\frac{x}{n}\right)^{n+1}$ decreases to $e^x$

so for an arbitrary but fixed $n$,

$$e^x \leq \left(1+\frac{x}{n}\right)^{n+1}$$

And this gives $$\frac{e^x-1}{x} \leq \frac{1}{x}\left[\left(1+\frac{x}{n}\right)^{n+1}-1\right]$$ Now take the limit $x\to 0$ we get

$$1\leq \lim\limits_{x\to 0} \frac{e^x-1}{x} \leq \frac{n+1}{n}$$

Since $n$ was arbitrary the result follows.