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# It is given that if $G\left( x \right)=\sqrt{25-{{x}^{2}}}$ , then the value of the following limit expression: $\underset{x\to 1}{\mathop{\lim }}\,\dfrac{G\left( x \right)-G\left( 1 \right)}{x-1}$ is equal to:(a) $-\dfrac{1}{2\sqrt{6}}$ (b) $\dfrac{1}{5}$ (c) $-\dfrac{1}{\sqrt{6}}$ (d) $\dfrac{1}{\sqrt{6}}$

Last updated date: 25th Feb 2024
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Hint: By substituting the value of $x=1$ in the limit expression we get $\dfrac{0}{0}$ form which is an indeterminate form so we apply L'Hopital's rule to find the value of limit in which we differentiate numerator and denominator with respect to x and then substitute the value of $x=1$ in the differentiated expression. In this way, we will get the value of the limit.

In the above problem, we have given:
$G\left( x \right)=\sqrt{25-{{x}^{2}}}$
The limit that we have to evaluate in the above problem is:
$\underset{x\to 1}{\mathop{\lim }}\,\dfrac{G\left( x \right)-G\left( 1 \right)}{x-1}$
Substituting the value of $G\left( x \right)=\sqrt{25-{{x}^{2}}}$ in the above limit expression we get,
$\underset{x\to 1}{\mathop{\lim }}\,\dfrac{\sqrt{25-{{x}^{2}}}-G\left( 1 \right)}{x-1}$
Finding the value of G(1) by substituting the value of $x=1$ in $G\left( x \right)=\sqrt{25-{{x}^{2}}}$ we get,
\begin{align} & G\left( 1 \right)=\sqrt{25-{{\left( 1 \right)}^{2}}} \\ & \Rightarrow G\left( 1 \right)=\sqrt{24} \\ \end{align}
Now, substituting $G\left( 1 \right)=\sqrt{24}$ in the limit expression we get,
$\underset{x\to 1}{\mathop{\lim }}\,\dfrac{\sqrt{25-{{x}^{2}}}-\sqrt{24}}{x-1}$
Applying limit on the expression written inside the limit by substituting $x=1$ we get,
\begin{align} & \dfrac{\sqrt{25-{{\left( 1 \right)}^{2}}}-\sqrt{24}}{1-1} \\ & =\dfrac{\sqrt{24}-\sqrt{24}}{0} \\ & =\dfrac{0}{0} \\ \end{align}
As you can see that the value of limit is in the form of $\dfrac{0}{0}$ which is an indeterminate form so we are going to apply L’Hospital’s rule to find the value of limit. In L'Hospital's rule we differentiate numerator and denominator with respect to x and then substitute the value of $x=1$ .
Let us assume $f\left( x \right)=\sqrt{25-{{x}^{2}}}-\sqrt{24}$ and $g\left( x \right)=x-1$ and then applying L’Hospital’s rule we get,
$\underset{x\to 1}{\mathop{\lim }}\,\dfrac{f(x)}{g(x)}=\underset{x\to 1}{\mathop{\lim }}\,\dfrac{f'(x)}{g'(x)}$ …………. Eq. (1)
Now, solving f’(x) and g’(x) we get,
$f'\left( x \right)=\dfrac{-\left( 2x \right)}{2\sqrt{25-{{x}^{2}}}}$
2 will be cancelled out from numerator and denominator we get,
$f'\left( x \right)=\dfrac{-\left( x \right)}{\sqrt{25-{{x}^{2}}}}$
$g'\left( x \right)=1$
Substituting the value of $f'\left( x \right)\And g'\left( x \right)$ in eq. (1) we get,
\begin{align} & \underset{x\to 1}{\mathop{\lim }}\,\dfrac{f'(x)}{g'(x)} \\ & =\underset{x\to 1}{\mathop{\lim }}\,\dfrac{-x}{\sqrt{25-{{x}^{2}}}} \\ \end{align}
Substituting $x=1$ in the above limit expression we get,
\begin{align} & \dfrac{-\left( 1 \right)}{\sqrt{25-{{\left( 1 \right)}^{2}}}} \\ & =-\dfrac{1}{\sqrt{24}} \\ \end{align}
We can write 24 which is in the square root as $6\times 4$ we get,
\begin{align} & -\dfrac{1}{\sqrt{4\left( 6 \right)}} \\ & =-\dfrac{1}{2\sqrt{\left( 6 \right)}} \\ \end{align}
From the above calculations, we have evaluated the limit as $-\dfrac{1}{2\sqrt{\left( 6 \right)}}$ .
So, the correct answer is “Option A”.

Note: The point to be noted here is that while applying L’Hospital’s rule in the above solution, we have differentiated numerator and denominator just once and then substitute the value of $x=1$ and got the value of limit but in some questions after differentiating the numerator and denominator and then we put the limit in the differentiated expression again we are getting $\dfrac{0}{0}$ form so we have to again differentiate numerator and denominator of the first differentiation result and then substitute the limit. We have to do this differentiation till we have got the determinant value.