JEE Advanced 2022 Paper 1 · Q02 · Limits & Continuity
Let $\alpha$ be a positive real number. Let $f:\mathbb{R}\to\mathbb{R}$ and $g:(\alpha,\infty)\to\mathbb{R}$ be the functions defined by $f(x)=\sin\!\left(\dfrac{\pi x}{12}\right)$ and $g(x)=\dfrac{2\log_{e}(\sqrt{x}-\sqrt{\alpha})}{\log_{e}(e^{\sqrt{x}}-e^{\sqrt{\alpha}})}$. Then the value of $\displaystyle\lim_{x\to\alpha^{+}} f(g(x))$ is __________.
Reveal answer + step-by-step solution
Correct answer:0.5
Solution
As $x\to\alpha^{+}$, let $t=\sqrt{x}-\sqrt{\alpha}\to 0^{+}$. Numerator: $2\log_{e}t$. Denominator: $\log_{e}(e^{\sqrt{x}}-e^{\sqrt{\alpha}})=\log_{e}\!\left(e^{\sqrt{\alpha}}(e^{t}-1)\right)=\sqrt{\alpha}+\log_{e}(e^{t}-1)$. Since $e^{t}-1\sim t$, $\log_{e}(e^{t}-1)\sim\log_{e}t$, which $\to-\infty$. So denominator $\sim\log_{e}t$. Hence $g(x)\to\dfrac{2\log_{e}t}{\log_{e}t}=2$. Therefore $\lim f(g(x))=\sin(2\pi/12)=\sin(\pi/6)=1/2=0.5$.
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