Question

If $G\left( x \right)=-\sqrt{25-{{x}^{2}}}$, then $\underset{x\to 1}{\mathop{\lim }}\,\dfrac{G\left( x \right)-G\left( 1 \right)}{x-1}$ is
A. $\dfrac{1}{\sqrt{24}}$
B. $\dfrac{1}{5}$
C. $-\sqrt{24}$
D. None of these

Answer 1

In the given question, we have been asked to find the limit of a given expression and we have to solve the given question using the L-hospital rule. In order to solve the question, first we need to differentiate the numerator and denominator with respect to ‘x’ one by one. Then by taking the limit x = 1 as it is given in the question we will put x = 1 in the differential expression. Then simplified further to get the answer.

Complete step by step solution:
We have given that,
$G\left( x \right)=-\sqrt{25-{{x}^{2}}}$
And
As we know that,
$\underset{x\to 1}{\mathop{\lim }}\,\dfrac{G\left( x \right)-G\left( 1 \right)}{x-1}$ is of $\dfrac{0}{0}$ form.
Since when we put the value of x = 1 the numerator and denominator become zero.
Thus,
Solving $\underset{x\to 1}{\mathop{\lim }}\,\dfrac{G\left( x \right)-G\left( 1 \right)}{x-1}$, using the L-hospital rule.
Here,
Differentiate numerator and denominator one-by-one,
$\dfrac{d}{dx}G\left( x \right)-G\left( 1 \right)=G'\left( x \right)-0$, and
$\dfrac{d}{dx}x-1=1-0$
Therefore,
$\dfrac{d}{dx}\left( \dfrac{G\left( x \right)-G\left( 1 \right)}{x-1} \right)=\dfrac{G'\left( x \right)-0}{1-0}=G'\left( x \right)$
Now,
Taking $G'\left( x \right)$
We have,
$G\left( x \right)=-\sqrt{25-{{x}^{2}}}$
Differentiate the above function using the chain rule, we get
$G'\left( x \right)=-\dfrac{1}{2\sqrt{25-{{x}^{2}}}}\times -2x=\dfrac{x}{\sqrt{25-{{x}^{2}}}}$
Thus,
$G'\left( x \right)=\dfrac{x}{\sqrt{25-{{x}^{2}}}}$
Now taking the limit ‘x’ approaches to 1,
$G'\left( 1 \right)=\dfrac{1}{\sqrt{25-{{1}^{2}}}}=\dfrac{1}{\sqrt{24}}$
Therefore,
$\underset{x\to 1}{\mathop{\lim }}\,\dfrac{G\left( x \right)-G\left( 1 \right)}{x-1}=\dfrac{1}{\sqrt{24}}$

So, the correct answer is Option A.

Note: While solving these types of problems, students need to be very careful while doing the calculation part to avoid making any type of error. They need to know about the concept of the finding the value of limit of a given expression using l’hospital rule. Instead of using l’hospital rule we can factorizing the polynomial and then putting the given value of x i.e. the limit and simplified further. Both the ways we will get the same answer.