Question

If \[f\left( 2 \right) = 4\], and \[f'\left( 2 \right) = 4\]. Then what is the value of \[\mathop {lim}\limits_{x \to 2} \dfrac{{\left[ {xf\left( 2 \right) - 2f\left( x \right)} \right]}}{{\left( {x - 2} \right)}}\]?
A. 2
B. \[ - 2\]
C. \[ - 4\]
D. 3

Answer 1

Hint: Simplify the given limit by using the L’Hospital rule. Then 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 functions are \[f\left( 2 \right) = 4\], and \[f'\left( 2 \right) = 4\].
Let’s solve the given limit.
Let \[L\] be the value of the limit.
\[L = \mathop {lim}\limits_{x \to 2} \dfrac{{xf\left( 2 \right) - 2f\left( x \right)}}{{\left( {x - 2} \right)}}\]
Substitute \[f\left( 2 \right) = 4\] in above equation.
\[L = \mathop {lim}\limits_{x \to 2} \dfrac{{4x - 2f\left( x \right)}}{{\left( {x - 2} \right)}}\]
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 \[x\].
\[L = \mathop {lim}\limits_{x \to 2} \dfrac{{4 - 2f'\left( x \right)}}{{1 - 0}}\]
\[ \Rightarrow \]\[L = \mathop {lim}\limits_{x \to 2} \left[ {4 - 2f'\left( x \right)} \right]\]
Simplify the above limits.
\[L = 4 - 2f'\left( 2 \right)\]
Substitute \[f'\left( 2 \right) = 4\] in above equation.
\[L = 4 - 2\left( 4 \right)\]
\[ \Rightarrow \]\[L = 4 - 8\]
\[ \Rightarrow \]\[L = - 4\]
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.