Question

How do you solve the following limit $ \dfrac{{{e^x} - 1}}{x} $ as x approaches zero?

Answer 1

Hint: The derivative of a function is a measure to find out the instantaneous change in a function. An equation containing one or more derivatives is called a differential equation. So we use L’Hospital’s rule in this question, the L’Hospital’s rule helps us to find the limits of an indifferential/indeterminate equation. This rule converts an indeterminate equation to a form containing differentials in both numerator and denominator that converts it to a limit that can be easily evaluated. Thus using L’Hospital’s rule we can find out the given limit.

Complete step-by-step answer:
We have to find the limit of $ \dfrac{{{e^x} - 1}}{x} $ as x approaches zero, so we apply L’Hospital’s rule according to which –
  $
  \mathop {\lim }\limits_{x \to c} \dfrac{{f(x)}}{{g(x)}} = \mathop {\lim }\limits_{x \to c} \dfrac{{f'(x)}}{{g'(x)}} \\
   \Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{{e^x} - 1}}{x} = \mathop {\lim }\limits_{x \to 0} \dfrac{{\dfrac{d}{{dx}}({e^x} - 1)}}{{\dfrac{{dx}}{{dx}}}} \\
   \Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{{e^x} - 1}}{x} = \mathop {\lim }\limits_{x \to 0} \dfrac{{{e^x}}}{1} = {e^0} \\
   \Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{{e^x} - 1}}{x} = 1 \\
   $
Hence the limit of $ \dfrac{{{e^x} - 1}}{x} $ as x approaches zero is equal to 1.
So, the correct answer is “1”.

Note: Now, according to L’Hospital’s rule $ \mathop {\lim }\limits_{x \to c} \dfrac{{f(x)}}{{g(x)}} = \mathop {\lim }\limits_{x \to c} \dfrac{{f'(x)}}{{g'(x)}} $ but there are certain conditions for this relation. The conditions are that the individual limits of the functions $ f(x) $ and $ g(x) $ should be equal to zero or infinite that is $ \mathop {\lim }\limits_{x \to c} f(x) = \mathop {\lim }\limits_{x \to c} g(x) = 0\,or\, \pm \infty $ and the differential of the function in the denominator should not be equal to zero that is $ g'(x) \ne 0 $ and $ \mathop {\lim }\limits_{x \to c} \dfrac{{f'(x)}}{{g'(x)}} $ must exist.
On putting $ x = 0 $ to find the limit of $ \dfrac{{{e^x} - 1}}{x} $ , we see that the limit doesn’t exist, we also see that $ \mathop {\lim }\limits_{x \to 0} {e^x} - 1 = {e^0} - 1 = 1 - 1 = 0 $ and $ \mathop {\lim }\limits_{x \to 0} x = 0 $ that is $ \mathop {\lim }\limits_{x \to 0} {e^x} - 1 = \mathop {\lim }\limits_{x \to 0} x = 0 $ . Now, $
  g'(x) = \dfrac{{dx}}{{dx}} = 1 \\
   \Rightarrow g'(x) \ne 0 \\
   $ and as we saw in the above solution $ \mathop {\lim }\limits_{x \to 0} \dfrac{{{e^x}}}{1} = \mathop {\lim }\limits_{x \to 0} {e^x} $ exists, thus the limit in the given question satisfies all the conditions of L’Hospital’s rule.