Question

If $ f\left( 9 \right)=9 $ , $ {{f}^{'}}\left( 9 \right)=4 $ , then find the value of given limit: $ \displaystyle \lim_{x \to 9}\dfrac{\sqrt{f\left( x \right)}-3}{\sqrt{x}-3} $ ?
(a) 4
(b) $ \dfrac{1}{4} $
(c) $ \dfrac{1}{2} $
(d) $ -\dfrac{1}{2} $

Answer 1

Hint: We start solving the problem by substituting the limit value in place of x which comes out to be indeterminate form. We then recall the definition of L-Hospital rule as “if $ \displaystyle \lim_{x \to a}\dfrac{g\left( x \right)}{h\left( x \right)} $ comes out to be indeterminate form $ \left( \dfrac{0}{0},\dfrac{\infty }{\infty },.... \right) $ , then the given limit can be found as $ \displaystyle \lim_{x \to a}\dfrac{g\left( x \right)}{h\left( x \right)}=\displaystyle \lim_{x \to a}\dfrac{{{g}^{'}}\left( x \right)}{{{h}^{'}}\left( x \right)}=\displaystyle \lim_{x \to a}\dfrac{{{g}^{''}}\left( x \right)}{{{h}^{''}}\left( x \right)}=...... $ ”. We then apply the L-Hospital rule to the given Limit and make the necessary calculations to reduce the given limits. We then substitute the limits and make use of the given values to get the required answer.

Complete step by step answer:
According to the problem, we are given that $ f\left( 9 \right)=9 $ , $ {{f}^{'}}\left( 9 \right)=4 $ and we need to find the value of the limit $ \displaystyle \lim_{x \to 9}\dfrac{\sqrt{f\left( x \right)}-3}{\sqrt{x}-3} $ .
Let us assume $ L=\displaystyle \lim_{x \to 9}\dfrac{\sqrt{f\left( x \right)}-3}{\sqrt{x}-3} $ .
 $ \Rightarrow L=\dfrac{\sqrt{f\left( 9 \right)}-3}{\sqrt{9}-3} $ .
 $ \Rightarrow L=\dfrac{\sqrt{9}-3}{3-3} $ .
 $ \Rightarrow L=\dfrac{3-3}{0} $ .
 $ \Rightarrow L=\dfrac{0}{0} $ , which is an indeterminate form.
So, we can make use of the L-Hospital rule.
Let us recall the L-Hospital rule. We know that if $ \displaystyle \lim_{x \to a}\dfrac{g\left( x \right)}{h\left( x \right)} $ comes out to be indeterminate form $ \left( \dfrac{0}{0},\dfrac{\infty }{\infty },.... \right) $ , then the given limit can be found as $ \displaystyle \lim_{x \to a}\dfrac{g\left( x \right)}{h\left( x \right)}=\displaystyle \lim_{x \to a}\dfrac{{{g}^{'}}\left( x \right)}{{{h}^{'}}\left( x \right)}=\displaystyle \lim_{x \to a}\dfrac{{{g}^{''}}\left( x \right)}{{{h}^{''}}\left( x \right)}=...... $ .
Now, let us apply the L-Hospital rule to the given limit.
So, we have $ L=\displaystyle \lim_{x \to 9}\dfrac{\dfrac{d}{dx}\left( \sqrt{f\left( x \right)}-3 \right)}{\dfrac{d}{dx}\left( \sqrt{x}-3 \right)} $ .
\[\Rightarrow L=\displaystyle \lim_{x \to 9}\dfrac{\dfrac{d}{dx}\left( \sqrt{f\left( x \right)} \right)-\dfrac{d}{dx}\left( 3 \right)}{\dfrac{d}{dx}\left( \sqrt{x} \right)-\dfrac{d}{dx}\left( 3 \right)}\].
We know that $ \dfrac{d}{dx}\left( \sqrt{x} \right)=\dfrac{1}{2\sqrt{x}} $ and $ \dfrac{dc}{dx}=0 $ , where c is a constant.
\[\Rightarrow L=\displaystyle \lim_{x \to 9}\dfrac{\dfrac{1}{2\sqrt{f\left( x \right)}}\dfrac{d}{dx}\left( f\left( x \right) \right)-0}{\dfrac{1}{2\sqrt{x}}-0}\].
\[\Rightarrow L=\displaystyle \lim_{x \to 9}\dfrac{\dfrac{{{f}^{'}}\left( x \right)}{2\sqrt{f\left( x \right)}}}{\dfrac{1}{2\sqrt{x}}}\].
\[\Rightarrow L=\dfrac{\dfrac{{{f}^{'}}\left( 9 \right)}{2\sqrt{f\left( 9 \right)}}}{\dfrac{1}{2\sqrt{9}}}\].
\[\Rightarrow L=\dfrac{\dfrac{4}{\sqrt{9}}}{\dfrac{1}{3}}\].
\[\Rightarrow L=\dfrac{\dfrac{4}{3}}{\dfrac{1}{3}}\].
\[\Rightarrow L=4\].
So, we have found the value of the given limit as 4.
 $ \, therefore, $ The correct option for the given problem is (a).

Note:
 Whenever we get the limit in indeterminate forms like $ \dfrac{0}{0},\dfrac{\infty }{\infty } $, we should make use of the L-Hospital rule. We should not use the L-Hospital rule if the applied limit doesn’t come out as an indeterminate form which is the most common mistake done by students. We should confuse or make mistakes while making calculations involving the differentiation of terms. We should not differentiate the whole limit as a single function, we should differentiate both numerator and denominator separately which should be in mind while solving this problem.