Question

Find the value of \[\mathop {\lim }\limits_{x \to 0} \dfrac{{\left[ {{e^x} - {e^{sinx}}} \right]}}{{\left[ {x - \sin x} \right]}} = \]
A. \[ - 1\]
B. \[0\]
C. \[1\]
D. None of these

Answer 1

Hint: We use the L’ Hospital rule to solve this question. L’ Hospital rule states that if \[\dfrac{{f\left( x \right)}}{{g\left( x \right)}}\] is in the form \[\dfrac{0}{0}\] or \[\dfrac{\infty }{\infty }\] when \[x = a\] plugs in, then \[\mathop {\lim }\limits_{x \to a} \dfrac{{f\left( x \right)}}{{g\left( x \right)}} = \mathop {\lim }\limits_{x \to a} \dfrac{{f'\left( x \right)}}{{g'\left( x \right)}}\]
To compute the limit, all we have to do is take the derivative of the numerator as well as the denominator.

Formula used:
We have been using the following formula:
1. \[\mathop {\lim }\limits_{x \to a} \dfrac{{f\left( x \right)}}{{g\left( x \right)}} = \mathop {\lim }\limits_{x \to a} \dfrac{{f'\left( x \right)}}{{g'\left( x \right)}}\]
2. \[\dfrac{d}{{dx}}\left( {\sin x} \right) = \cos x\]
3. \[\dfrac{d}{{dx}}\left( {u \cdot v} \right) = u \cdot \dfrac{{dv}}{{dx}} + v \cdot \dfrac{{du}}{{dx}}\]
Complete step-by-step solution:
Now we have been given that \[\mathop {\lim }\limits_{x \to 0} \dfrac{{\left[ {{e^x} - {e^{sinx}}} \right]}}{{\left[ {x - \sin x} \right]}}\]
Now, by applying the L'Hospital rule, as the given problem is in indeterminate form
\[\mathop {\lim }\limits_{x \to 0} \dfrac{{\left[ {{e^x} - {e^{sinx}}} \right]}}{{\left[ {x - \sin x} \right]}} = \mathop {\lim }\limits_{x \to 0} \dfrac{{\left[ {{e^x} - {e^{sinx}}} \right]'}}{{\left[ {x - \sin x} \right]'}}\]
By taking a derivative, we get
\[
  \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{{e^x} - {e^{\sin x}}}}{{x - \sin x}}} \right) = \mathop {\lim }\limits_{x \to 0} \dfrac{{{e^x} - {e^{\sin x}}\cos x}}{{1 - \cos x}} \\
   = \mathop {\lim }\limits_{x \to 0} \dfrac{{{e^x} - \left( {{e^{\sin x}}\left( { - \sin x} \right) + {e^{sinx}}\cos x\cos x} \right)}}{{0 - \left( { - \sin x} \right)}} \\
   = \mathop {\lim }\limits_{x \to 0} \dfrac{{{e^x} + {e^{\sin x}} \cdot \sin x - {{\cos }^2}x{e^{\sin x}}}}{{\sin x}} \\
   = \mathop {\lim }\limits_{x \to 0} \dfrac{{{e^x} + {e^{\sin x}}\cos x + \sin x \cdot {e^{\sin x}} - {{\cos }^2}x{e^{\sin x}}{{cos x - 2\cos x\left( { - \sin x} \right){e^{\sin x}}}}}}{{\cos x}} \\
 \]
By applying the limit, we get
\[
  \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{{e^x} - {e^{\sin x}}}}{{x - \sin x}}} \right) = \dfrac{{{e^0} + {e^0} \times 1 + 0 \cdot {e^0} \cdot 1 - {1^2} \cdot {e^0} \cdot 1 - 2 \cdot 1\left( { - 0} \right){e^0}}}{1} \\
   = 1 + 1 - 1\left( {\because {e^0} = 1} \right) \\
   = 1 \\
 \]

Hence, option (C) is the correct option

Additional information:When computing sums on limits, a stalemate can occur when the numerator and denominator of the limit produce a \[\dfrac{0}{0}\] or \[\dfrac{\infty }{\infty }\]. An indeterminate form is what it's termed. In this scenario, a way out is to continue computing the limit by differentiating both the numerator and the denominator until the numerator and denominator no longer provide the indeterminate form. The L-Hospital rule is the method of taking derivatives of the numerator and denominator of a limit and it can be used many times.

Note: Students must remember the condition of the L'Hospital rule, which states that a fraction must be of two functions and the function must be indeterminate, and they can only use this method directly if the limit problem is not indeterminate.