Let $\alpha$ be a positive real number. Let $f: \mathbb{R} \rightarrow \mathbb{R}$ and $g:(\alpha, \infty) \rightarrow \mathbb{R}$ be the functions defined by $$f(x)=\sin \left(\frac{\pi x}{12}\right) \text { and } g(x)=\frac{2 \log _{e}(\sqrt{x}-\sqrt{\alpha})}{\log _{e}\left(e^{\sqrt{x}}-e^{\sqrt{\alpha}}\right)}.$$ Then the value of $\lim _{x \rightarrow \alpha^{+}} f(g(x))$ is ____.

JEE Advanced · 2022 · 28 Aug 2022 · Paper 1 · Limits XII

Let

α

be a positive real number. Let

f : R \to R

and

g : (α, \infty) \to R

be the functions defined by

f (x) = sin (\frac{π x}{12}) and g (x) = \frac{2 lo g _{e} ( x - α )}{lo g _{e} ( e ^{x} - e ^{α} )} .

Then the value of

lim_{x \to α^{+}} f (g (x))

is ____.

Answer: 0.5

We need to find

lim_{x \to α^{+}} g (x)

first. This is a

\frac{0}{0}

form, so applying L'Hôpital's rule:

x \to α^{+} lim \frac{2 ln ( x - α )}{ln ( e ^{x} - e ^{α} )} = 2 x \to α^{+} lim \frac{\frac{1}{x - α} \cdot \frac{1}{2 x}}{\frac{e ^{x}}{e ^{x} - e ^{α}} \cdot \frac{1}{2 x}}

= \frac{2}{e ^{α}} x \to α^{+} lim \frac{e ^{x} - e ^{α}}{x - α}

Applying L'Hôpital's rule again:

= \frac{2}{e ^{α}} x \to α^{+} lim \frac{e ^{x} \cdot \frac{1}{2 x}}{\frac{1}{2 x}} = \frac{2}{e ^{α}} \cdot e^{α} = 2

Therefore,

lim_{x \to α^{+}} f (g (x)) = f (2) = sin \frac{π \cdot 2}{12} = sin \frac{π}{6} = \frac{1}{2} = 0.50

JEE Advanced · 2022 · 28 Aug 2022 · Paper 1 · Numerical answer