Question

Find the value of $ \mathop {\lim }\limits_{x \to 0} \dfrac{{{3^{2 + x}} - 9}}{x} $ .
A) $ \log {}_e3 $
B) $ 9\log {}_e3 $
C) $ 9\log {}_3e $
D) $ 3\log {}_3e $

Answer 1

Hint: Here when we directly substitute the limit value x=0, we get 0/0 which is not defined. So we have to use L’hospital’s Rule to solve the above limit. L'Hospital's Rule states that if we have an indeterminate form 0/0, all we need to do is differentiate the numerator and the denominator and then take the limit.

Complete step-by-step answer:
We are given to find the value of $ \mathop {\lim }\limits_{x \to 0} \dfrac{{{3^{2 + x}} - 9}}{x} $
Here the limit of x tends to zero.
By direct substitution, substitute the value x=0 in the limit as it tends to the value zero.
 $
  \mathop {\lim }\limits_{x \to 0} \dfrac{{{3^{2 + x}} - 9}}{x} \\
  x = 0 \\
   = \dfrac{{{3^{2 + 0}} - 9}}{0} \\
   = \dfrac{{9 - 9}}{0} \\
   = \dfrac{0}{0} \\
  $
The value of 0/0 is undefined. So use L’Hospital’s rule to solve the limit.
L’Hospital’s Rule states that If $ \mathop {\lim }\limits_{x \to a} \dfrac{{f\left( x \right)}}{{g\left( x \right)}} = \dfrac{0}{0} $ then $ \mathop {\lim }\limits_{x \to a} \dfrac{{f\left( x \right)}}{{g\left( x \right)}} = \mathop {\lim }\limits_{x \to a} \dfrac{{\dfrac{d}{{dx}}\left[ {f\left( x \right)} \right]}}{{\dfrac{d}{{dx}}\left[ {g\left( x \right)} \right]}} $
 $
  f\left( x \right) = {3^{2 + x}} - 9 \\
  \dfrac{d}{{dx}}\left[ {f\left( x \right)} \right] = \dfrac{d}{{dx}}\left( {{3^{2 + x}} - 9} \right) \\
   = \dfrac{d}{{dx}}\left( {{3^{2 + x}}} \right) - \dfrac{d}{{dx}}\left( 9 \right) \\
   = {\log _e}3\left( {{3^{2 + x}}} \right) - 0 \\
  $
Differentiation of a constant is zero.
Differentiation of
  $
  {a^x} = \ln a\left( {{a^x}} \right) \\
  \ln a = {\log _e}a \\
  \dfrac{d}{{dx}}\left[ {f\left( x \right)} \right] = {\log _e}3\left( {{3^{2 + x}}} \right) \\
  g\left( x \right) = x \\
  \dfrac{d}{{dx}}\left[ {g\left( x \right)} \right] = \dfrac{d}{{dx}}\left( x \right) \\
  \dfrac{d}{{dx}}\left[ {g\left( x \right)} \right] = 1 \\
  \mathop {\lim }\limits_{x \to 0} \dfrac{{f\left( x \right)}}{{g\left( x \right)}} = \mathop {\lim }\limits_{x \to 0} \dfrac{{{{\log }_e}3\left( {{3^{2 + x}}} \right)}}{1} = \mathop {\lim }\limits_{x \to 0} {\log _e}3\left( {{3^{2 + x}}} \right) \\
  \mathop {\lim }\limits_{x \to 0} {\log _e}3\left( {{3^{2 + x}}} \right) = {\log _e}3\left( {{3^{2 + 0}}} \right) \\
   = {\log _e}3\left( {{3^2}} \right) \\
   = 9{\log _e}3 \\
  $
Therefore, the value of $ \mathop {\lim }\limits_{x \to 0} \dfrac{{{3^{2 + x}} - 9}}{x} $ is $ 9{\log _e}3 $
So, the correct answer is “Option B”.

Note: A limit is the value that a function approaches as the input approaches some value. Limits are essential to calculus and mathematical analysis, derivatives and integrals. The limit of a constant function is equal to the constant. Limits follow distributive property in addition and subtraction. Do not use L’Hospital’s Rule when you have a determinable form of limit.