Question 2 - Jee advanced Math 2022 P1 Questions with Solutions

2) Let \(\alpha\) be a positive real number. Let \(f:R \rightarrow R\) and \(g:(\alpha, \infty) \rightarrow R\) be the functions defined by
\(f(x) = \frac{\sin{\pi x}}{12}\)  and \(g(x) = \frac{2\ln{(\sqrt{x}-\sqrt{\alpha}})}{\ln{(e^{\sqrt{x}}-e^{\sqrt{\alpha}})}}\).

Then the value of \(\lim_{x\rightarrow \alpha^{+}}{f(g(x))}\) is ______.

Sol : 

Limit of a composite function \(f(g(x))\) as \(x\) approaches \(a\) can be found using the following result:
If,       i) \(\lim_{x\rightarrow a}{g(x)}\) exists and is equal to L
and     ii) f(x) is continuous at L
then \(\lim_{x\rightarrow a}{f(g(x))} = f(\lim_{x\rightarrow a}{g(x)}) = f(L)\)

i)  Let’s check the first condition for our example. 

\(\lim_{x\rightarrow \alpha^{+}}{g(x)} = \lim_{x\rightarrow \alpha^{+}}{\frac{2\ln{(\sqrt{x}-\sqrt{\alpha})}}{\ln{(e^{\sqrt{x}}-e^{\sqrt{\alpha}})}}}\)

On substituting \(x = \alpha\) the limit will result in indeterminate \(\frac{0}{0}\) form. So L'Hôpital's rule can be used to evaluate the limit. This rule says that 
\(\lim_{x \rightarrow \alpha^{+}}{\frac{g_{1}(x)}{g_{2}(x)}} = \lim_{x \rightarrow \alpha^{+}}{\frac{g_{1}^{’}(x)}{g_{2}^{’}(x)}}\) if the following conditions are satisfied.

{1} \(\lim_{x \rightarrow p}{\frac{g_{1}(x)}{g_{2}(x)}}\) yields 0/0 form. 
which is indeed the case in our example too.

{2} \(g_{1}^{’}(x)\) and \(g_{2}^{’}(x)\) are differentiable functions everywhere in the domain except possible at \(x = p\). 

\(g_{1}^{’}(x) = 2\frac{1}{\sqrt{x}-\sqrt{\alpha}} \times \frac{1}{2\sqrt{x}} \times {1}\)

                 \( = \frac{1}{\sqrt{x}-\sqrt{\alpha}} \times \frac{1}{\sqrt{x}}\)

It is differentiable at every real value of \(x\) except at \(x=0, \alpha\). However, \((\alpha, \infty)\) do not contain either of the two values. So \(g_{1}^{’}(x)\) is differentiate in the given domain. 

Similarly, we will get

\(g_{2}{’}(x) = \frac{1}{e^{\sqrt{x}}-e^{\sqrt{\alpha}}} \times e^\sqrt{x} \times \frac{1}{2\sqrt{x}}\)
which is also differentiable everywhere in the interval \((\alpha, \infty)\).

{3} the denominator \(g_{2}^{’}(x) \neq 0\) for every \(x\) in the domain except possibly at \(x = p\)
….which is also satisfied by our example

{4} \(\lim_{x \rightarrow p}{\frac{g_{1}^{’}(x)}{g_{2}^{’}(x)}}\) exists

 \(\lim_{x \rightarrow \alpha^{+}}{\frac{g_{1}^{’}(x)}{g_{2}^{’}(x)}} = \lim_{x \rightarrow \alpha^{+}}{\frac{2(e^{\sqrt{x}}-e^{\sqrt{\alpha}})}{e^{\sqrt{x}}\times (\sqrt{x}-\sqrt{\alpha})}}\)(after simplification)

               \(= \lim_{x \rightarrow \alpha^{+}}{\frac{2}{e^{\sqrt{x}}}} \times \lim_{x \rightarrow \alpha^{+}}{\frac{e^{\sqrt{x}}-e^{\sqrt{\alpha}}}{\sqrt{x}-\sqrt{\alpha}}}\)

For the second limit we will use the result 
\(\lim_{t \rightarrow 0}{\frac{e^{t} - 1}{t}} = 1\)

Let \(\sqrt{x} - \sqrt{\alpha} = t\)
\(x \rightarrow \alpha^{+} \implies t \rightarrow 0^{+}\)
(As x approaches \(\alpha\) from right hand side, t approaches 0 from right hand side)

\(\lim_{x \rightarrow \alpha^{+}}{\frac{e^{\sqrt{x}}-e^{\sqrt{\alpha}}}{\sqrt{x}-\sqrt{\alpha}}}\)

\(= lim_{x \rightarrow \alpha^{+}}{\frac{e^{\sqrt{\alpha}}(e^{\sqrt{x}-\sqrt{\alpha}} - 1)}{\sqrt{x} - \sqrt{\alpha}}}\)

\(= e^{\sqrt{\alpha}} lim_{t \rightarrow 0^{+}}{\frac{e^{t}-1}{t}}\)

\( = e^{\sqrt{\alpha}} \times 1\)

\( = e^{\sqrt{\alpha}}\)

Therefore, 
 \( \lim_{x \rightarrow \alpha^{+}}{\frac{g_{1}^{’}(x)}{g_{2}^{’}(x)}} = \lim_{x \rightarrow \alpha^{+}}{\frac{2}{e^{\sqrt{x}}}} \times e^{\sqrt{\alpha}}\)

          \(= \frac{2}{e^{\sqrt{\alpha}}} \times e^{\sqrt{\alpha}}\)

           \( = 2\)

Therefore,  \( \lim_{x \rightarrow \alpha^{+}}{\frac{g_{1}^{’}(x)}{g_{2}^{’}(x)}}\) exists

All four conditions of L'Hôpital's rule are satisfied. 

\(\implies \lim_{x \rightarrow \alpha^{+}}{g(x)}\)
            
           \(= \lim_{x \rightarrow \alpha^{+}}{\frac{g_{1}(x)}{g_{2}(x)}}\)
           
           \(= \lim_{x \rightarrow \alpha^{+}}{\frac{g_{1}^{’}(x)}{g_{2}^{’}(x)}}\)
         
            \( = 2\)

\(\implies \lim_{x \rightarrow \alpha^{+}}{g(x)}\) exists

The first condition of the theorem at the beginning of this post is verified

ii) \(f(x) = \sin{\frac{\pi x}{12}}\) is a continuous function at every real number. So it is continuous at L=2.

\(\implies \lim_{x \rightarrow \alpha^{+}}{f(g(x))}\)

\(= f(\lim_{x\rightarrow \alpha^{+}}{g(x)})\)

      \( = f(2)\)
      \( = \sin{\frac{\pi (2)}{12}}\)
      \( = \sin{\frac{\pi}{6}}\)
      \( = \frac{1}{2}\) 
      \(= 0.5\)

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