Let \alpha be a positive real number. Let f: \mathbbR arrow \mathbbR and g:(\alpha, \infty) arrow \mathbbR be the functions defined by f(x)=\sin (\frac\pi x12) \quad and \quad g(x)=\frac2 \log _e(\sqrtx-\sqrt\alpha)\log _e(e^\sqrtx-e^\sqrt\alpha) . Then the value of \lim \limits_x arrow \alpha^+ f(g(x)) is

Limits Continuity and Differentiability PYQ — JEE Advanced 2022 · Shift 1 · Q20 with Solution

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Correct answer: 0.49TO0.51

\lim_{x\to \alpha^+} f(g(x)),\qquad f(x)=\sin\left(\frac{\pi x}{12}\right),\qquad g(x)=\frac{2\ln(\sqrt{x}-\sqrt{\alpha})}{\ln\left(e^{\sqrt{x}}-e^{\sqrt{\alpha}}\right)}.

Let $t=\sqrt{x}-\sqrt{\alpha}.$ As $x\to \alpha^+$ , we have $t\to 0^+$ . Also, $\sqrt{x}=\sqrt{\alpha}+t.$

Then $e^{\sqrt{x}}-e^{\sqrt{\alpha}}=e^{\sqrt{\alpha}+t}-e^{\sqrt{\alpha}}=e^{\sqrt{\alpha}}(e^t-1).$

So the denominator of $g(x)$ becomes

\ln\left(e^{\sqrt{x}}-e^{\sqrt{\alpha}}\right) =\ln\left(e^{\sqrt{\alpha}}(e^t-1)\right) =\sqrt{\alpha}+\ln(e^t-1).

Hence,

g(x)=\frac{2\ln t}{\sqrt{\alpha}+\ln(e^t-1)}.

So,

Since $\ln t\to -\infty$ , the constant $\sqrt{\alpha}$ and the $o(1)$ term are negligible compared to $\ln t$ . Thus,

\lim_{x\to \alpha^+} g(x)=\lim_{t\to 0^+}\frac{2\ln t}{\ln t+\sqrt{\alpha}+o(1)}=2.

\lim_{x\to \alpha^+} f(g(x))=f\left(\lim_{x\to \alpha^+} g(x)\right)=f(2).

Now,

f(2)=\sin\left(\frac{\pi\cdot 2}{12}\right)=\sin\left(\frac{\pi}{6}\right)=\frac12.

\boxed{\lim_{x\to \alpha^+} f(g(x))=\frac12}.

For integer-type range checking, this is $0.5$ .