Question

If \[f\left( 5 \right) = 7\], and \[f'\left( 5 \right) = 7\], then \[\mathop {\lim }\limits_{x \to 5} \dfrac{{xf\left( 5 \right) - 5f\left( x \right)}}{{x - 5}}\] is given by
A. 35
B. - 35
C. 28
D. - 28

Answer 1

Hint: In this question, we need to find the value \[\mathop {\lim }\limits_{x \to 5} \dfrac{{xf\left( 5 \right) - 5f\left( x \right)}}{{x - 5}}\]. For this, we have to use L-hospital’s rule. After that, we will put \[f\left( 5 \right) = 7\], and \[f'\left( 5 \right) = 7\] in the given limit and simplify it to get the desired result.

Complete step-by-step solution:
Given that \[\mathop {\lim }\limits_{x \to 5} \dfrac{{xf\left( 5 \right) - 5f\left( x \right)}}{{x - 5}}\]
Now, put \[x = 5\] in the above limit.
Thus, we get
\[\mathop {\lim }\limits_{x \to 5} \dfrac{{xf\left( 5 \right) - 5f\left( x \right)}}{{x - 5}} = \dfrac{{5f\left( 5 \right) - 5f\left( 5 \right)}}{{5 - 5}}\]
\[\mathop {\lim }\limits_{x \to 5} \dfrac{{xf\left( 5 \right) - 5f\left( x \right)}}{{x - 5}} = \dfrac{{5\left( 7 \right) - 5\left( 7 \right)}}{{5 - 5}}\]
By simplifying, we get
\[\mathop {\lim }\limits_{x \to 5} \dfrac{{xf\left( 5 \right) - 5f\left( x \right)}}{{x - 5}} = \dfrac{{35 - 35}}{{5 - 5}}\]
But it is in the form of \[\dfrac{0}{0}\].
So, to find the value of limit, we will use L hospitals rule here.
That means, we will take derivative of numerator and denominator separately and then put \[x = 5\] in the given limit.
Thus, we get
\[\mathop {\lim }\limits_{x \to 5} \dfrac{{xf\left( 5 \right) - 5f\left( x \right)}}{{x - 5}} = \mathop {\lim }\limits_{x \to 5} \dfrac{{\dfrac{d}{{dx}}\left( x \right)f\left( 5 \right) - 5\dfrac{d}{{dx}}f\left( x \right)}}{{\dfrac{d}{{dx}}\left( {x - 5} \right)}}\]
\[\mathop {\lim }\limits_{x \to 5} \dfrac{{xf\left( 5 \right) - 5f\left( x \right)}}{{x - 5}} = \mathop {\lim }\limits_{x \to 5} \dfrac{{f\left( 5 \right) - 5f'\left( x \right)}}{1}\]
But \[f\left( 5 \right) = 7\], and \[f'\left( 5 \right) = 7\]
So, we get
\[\mathop {\lim }\limits_{x \to 5} \dfrac{{xf\left( 5 \right) - 5f\left( x \right)}}{{x - 5}} = \dfrac{{7 - 5\left( 7 \right)}}{{\left( 1 \right)}}\]
\[\mathop {\lim }\limits_{x \to 5} \dfrac{{xf\left( 5 \right) - 5f\left( x \right)}}{{x - 5}} = 7 - 35\]
\[\mathop {\lim }\limits_{x \to 5} \dfrac{{xf\left( 5 \right) - 5f\left( x \right)}}{{x - 5}} = - 28\]
Hence, the value of \[\mathop {\lim }\limits_{x \to 5} \dfrac{{xf\left( 5 \right) - 5f\left( x \right)}}{{x - 5}}\] is \[ - 28\].
Therefore, the correct option is (D).

Additional information: L-hospital’s rule is properly called "lopeetals rule. "L-hospital’s rule is a particular way of evaluating undefined forms like \[\dfrac{0}{0}\] or \[\dfrac{\infty}{\infty} \]. L-hospital’s rule is often used in mathematics to analyze the limits of undefined forms for derivatives. The L-Hospital rule can be used multiple times. You can apply this rule again and again, and it will retain any undefined form after every application. L-hospital’s Rule cannot be applied if the problem is not in the undefined forms.

Note: Here, students generally make mistakes in applying the L-hospitals rule. Instead of taking the separate derivatives of numerator and denominator, they may apply the derivative to the whole. Due to this, they will get the wrong result.