Question

If \[f'\left( 3 \right) = 2\], then what is the value of \[\mathop {lim}\limits_{h \to 0} \left[ {\dfrac{{f\left( {3 + {h^2}} \right) - f\left( {3 - {h^2}} \right)}}{{2{h^2}}}} \right]\]?
A. 5
B. \[\dfrac{1}{5}\]
C. 2
D. None of these

Answer 1

Hint: Check the given limit. If the limit is in the form of \[\dfrac{0}{0}\] or \[\dfrac{\infty }{\infty }\], then simplify the given limit by using the L’Hospital rule. After that, substitute the given values of the functions in the limit and get the required answer.

Formula Used:
L’Hospital rule:
If \[\dfrac{{f\left( x \right)}}{{g\left( x \right)}}\] is in the form \[\dfrac{0}{0}\] or \[\dfrac{\infty }{\infty }\] when \[x \to c\], then \[\mathop {lim}\limits_{x \to c} \dfrac{{f\left( x \right)}}{{g\left( x \right)}} = \mathop {lim}\limits_{x \to c} \dfrac{{f'\left( x \right)}}{{g'\left( x \right)}}\].

Complete step by step solution:
The given limit is \[\mathop {lim}\limits_{h \to 0} \left[ {\dfrac{{f\left( {3 + {h^2}} \right) - f\left( {3 - {h^2}} \right)}}{{2{h^2}}}} \right]\].
Let’s solve the given limit.
Let \[L\] be the value of the limit.
\[L = \mathop {lim}\limits_{h \to 0} \left[ {\dfrac{{f\left( {3 + {h^2}} \right) - f\left( {3 - {h^2}} \right)}}{{2{h^2}}}} \right]\]
The right-hand side of the limit is the in form of \[\dfrac{0}{0}\].
So, apply the L’Hospital rule.
Differentiate the right-hand side of the limit with respect to \[h\].
\[L = \mathop {lim}\limits_{h \to 0} \left[ {\dfrac{{f'\left( {3 + {h^2}} \right) \times 2h - f'\left( {3 - {h^2}} \right) \times \left( { - 2h} \right)}}{{4h}}} \right]\]
Cancel out the common terms from the numerator and denominator.
\[L = \mathop {lim}\limits_{h \to 0} \left[ {\dfrac{{f'\left( {3 + {h^2}} \right) + f'\left( {3 - {h^2}} \right)}}{2}} \right]\]
\[ \Rightarrow \]\[L = \dfrac{1}{2}\mathop {lim}\limits_{h \to 0} \left[ {f'\left( {3 + {h^2}} \right) + f'\left( {3 - {h^2}} \right)} \right]\]
Now apply the limit.
\[L = \dfrac{1}{2}\left[ {f'\left( {3 + {0^2}} \right) + f'\left( {3 - {0^2}} \right)} \right]\]
Simplify the above equation.
\[L = \dfrac{1}{2}\left[ {f'\left( 3 \right) + f'\left( 3 \right)} \right]\]
\[ \Rightarrow \]\[L = \dfrac{1}{2}\left[ {2f'\left( 3 \right)} \right]\]
Substitute the value of \[f'\left( 3 \right)\] in the above equation.
\[L = \dfrac{1}{2}\left[ {2\left( 2 \right)} \right]\]
\[ \Rightarrow \]\[L = 2\]
Thus, the value of the limit is \[\mathop {lim}\limits_{h \to 0} \left[ {\dfrac{{f\left( {3 + {h^2}} \right) - f\left( {3 - {h^2}} \right)}}{{2{h^2}}}} \right] = 2\].

Hence the correct option is C.

Note: When the numerator and denominator of a limit are in the form of \[\dfrac{0}{0}\] or \[\dfrac{\infty }{\infty }\], we use L’Hospital rule. In this process, we take derivatives of the numerator and denominator with respect to the given variable. So that the function is no longer in the indeterminate form.