Question

How do you find the limit $\dfrac{{{e^x} - 1}}{x}$ as $x \to 0$ ?

Answer 1

Hint : In order to determine the limit of the above function, you can see by putting $x = 0$ the result is in the indeterminate form i.e. $\dfrac{0}{0}\,or\,\dfrac{{ \pm \infty }}{{ \pm \infty }}$ . To remove the indeterminate form use the L-hospital’s rule which says $\mathop {\lim }\limits_{x \to a} \dfrac{{f(x)}}{{g(x)}} = \mathop {\lim }\limits_{x \to a} \dfrac{{f'(x)}}{{g'(x)}}$ . In our case, consider $f(x) = {e^x} - 1$ and $g(x) = x$ and calculate their derivative with the help of derivative rules $\dfrac{d}{{dx}}\left( {{e^x}} \right) = {e^x}$ and $\dfrac{d}{{dx}}\left( c \right) = 0$ , put them into the L-hospital’s rule to obtain the limit of the function.

Complete step-by-step answer :
We are given a exponential function in variable $x$ i.e. $\dfrac{{{e^x} - 1}}{x}$ having limit $x \to 0$ .
$\mathop {\lim }\limits_{x \to 0} \dfrac{{{e^x} - 1}}{x}$
As you can see this is the limit problem, so if we directly put the limit $x = 0$ the result will have denominator and numerator both equal to zero i.e. of the indeterminate form $\dfrac{0}{0}$ which is completely not acceptable as it give the result as infinity.
So to avoid the indeterminate form, we will be using L-Hospital’s rule
According to L-Hospital’s rule if any limit has the following cases
$\mathop {\lim }\limits_{x \to a} \dfrac{{f(x)}}{{g(x)}} = \dfrac{0}{0}\,\,\,\,\,\,or\,\,\,\,\,\,\mathop {\lim }\limits_{x \to a} \dfrac{{f(x)}}{{g(x)}} = \dfrac{{ \pm \infty }}{{ \pm \infty }}$ where a is any real number.
So in such cases we calculate limit as $\mathop {\lim }\limits_{x \to a} \dfrac{{f(x)}}{{g(x)}} = \mathop {\lim }\limits_{x \to a} \dfrac{{f'(x)}}{{g'(x)}}$
In our case, we have $f(x) = {e^x} - 1$ and $g(x) = x$
Now calculating the derivative of $f\left( x \right)$ using the rule of derivative $\dfrac{d}{{dx}}\left( {{e^x}} \right) = {e^x}$ and $\dfrac{d}{{dx}}\left( c \right) = 0$
$f'\left( x \right) = {e^x}$ and
Derivative of $g\left( x \right)$ by using rule $\dfrac{d}{{dx}}\left( x \right) = 1$ , we get
$g'\left( x \right) = 1$
Putting these $f'\left( x \right)$ and $g'\left( x \right)$ in the L-Hospital rule , we get
$\mathop {\lim }\limits_{x \to a} \dfrac{{f(x)}}{{g(x)}} = \mathop {\lim }\limits_{x \to a} \dfrac{{f'(x)}}{{g'(x)}} \\\ \mathop {\lim }\limits_{x \to 0} \dfrac{{f(x)}}{{g(x)}} = \mathop {\lim }\limits_{x \to 0} \dfrac{{{e^x}}}{1} \\\ \mathop {\lim }\limits_{x \to 0} \dfrac{{f(x)}}{{g(x)}} = {e^0} \\\$
And as we know anything raised to the power is equal to one.
$\mathop {\lim }\limits_{x \to 0} \dfrac{{f(x)}}{{g(x)}} = 1$
$\mathop {\therefore \lim }\limits_{\,\,\,\,\,\,\,\,x \to 0} \dfrac{{{e^x} - 1}}{x} = 1$
Therefore, the limit of function $\dfrac{{{e^x} - 1}}{x}$ as $x \to 0$ is $1$ .
So, the correct answer is “1”.

Note : 1. Don’t forget to cross-check your answer.
2.After putting the Limit the result should never in the indeterminate form $\dfrac{0}{0}\,or\,\dfrac{{ \pm \infty }}{{ \pm \infty }}$ . If it is contained, apply some operation to modify the result or use the L-Hospitals rule .
3. $e$ is the exponential constant having value equal to $2.71828$ .