# limit of (e^x -1)/2x as x tends to 0?

Jan 16, 2018

${\lim}_{x \to 0} \frac{{e}^{x} - 1}{2 x} = \frac{1}{2}$

#### Explanation:

I'll assume you meant ${\lim}_{x \to 0} \frac{{e}^{x} - 1}{2 x}$

Investigate the limit first by using direct substitution:

${\lim}_{x \to 0} \frac{{e}^{x} - 1}{2 x} = \frac{{e}^{\textcolor{red}{0}} - 1}{2 \cdot \textcolor{red}{0}} = \frac{1 - 1}{0} = \frac{0}{0}$

We get the indeterminate form $\frac{0}{0}$ which should indicate that we should use L'hospital's Rule to rewrite the limit into something we can evaluate

L'Hospital states:

If ${\lim}_{x \to a} f \frac{x}{g \left(x\right)} = \frac{0}{0}$

Then, ${\lim}_{x \to a} f \frac{x}{g} \left(x\right) = {\lim}_{x \to a} \frac{\frac{d}{\mathrm{dx}} \left[f \left(x\right)\right]}{\frac{d}{\mathrm{dx}} \left[g \left(x\right)\right]}$

In this example:

$f \left(x\right) = {e}^{x} - 1 \implies \frac{d}{\mathrm{dx}} \left[{e}^{x} - 1\right] = {e}^{x}$

$g \left(x\right) = 2 x \implies \frac{d}{\mathrm{dx}} \left[2 x\right] = 2$

${\lim}_{x \to 0} \frac{{e}^{x} - 1}{2 x} = {\lim}_{x \to 0} \frac{\frac{d}{\mathrm{dx}} \left[{e}^{x} - 1\right]}{\frac{d}{\mathrm{dx}} \left[2 x\right]} = {\lim}_{x \to 0} {e}^{x} / 2$

Now we can attempt to evaluate the limit:

${\lim}_{x \to 0} {e}^{x} / 2 = {e}^{\textcolor{red}{0}} / 2 = \frac{1}{2}$

Jan 16, 2018

We can say it is around $\frac{1}{2}$

#### Explanation:

I am just trying to visualize what is going on here.

Let's graph the function $f \left(x\right) = \frac{{3}^{x} - 1}{2 x}$
graph{(3^x-1)/(2x) [-10, 10, -5, 5]}

We can see that the function is continuous except at $x = 0$.
We also can see that as $x$ approaches 0, it is getting really close to $\frac{1}{2}$ from both the positive and the negative directions.