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 $ .