Let a be a positive real number. Let f : and g : a , be the functions defined by f ( x ) = sin ( x 12 ) and g ( x ) = 2 log e ( x - ) log e ( e x - e ) . Then the value of lim x + f ( g ( x ) ) is ___________. [2022]

Question

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 $g$ $(x)$ $= \frac{2 \log_{e} (\sqrt{x} - \sqrt{α})}{\log_{e} (e^{\sqrt{x}} - e^{\sqrt{α}})} .$

Then the value of $\lim_{x \to α^{+}} f (g (x))$ is ___________. [2022]

Smart Achievers · Accepted Answer

(00.50)

$We have, g$ $(x)$ $= \frac{2 \log_{e} (\sqrt{x} - \sqrt{α})}{\log_{e} (e^{\sqrt{x}} - e^{\sqrt{α}})}$

Now, $\lim_{x \to α^{+}} g$ $(x)$ $= \lim_{x \to α^{+}} \frac{\frac{2}{\sqrt{x} - \sqrt{α}} (\frac{1}{2 \sqrt{x}})}{\frac{1}{e^{\sqrt{x}} - e^{\sqrt{α}}} (\frac{1}{2 \sqrt{x}} e^{\sqrt{x}})},$

${when x \to α^{+}, f (x) \to \frac{\infty}{\infty}}$

Apply L.H. Rule

$= \lim_{x \to α^{+}} \frac{e^{\sqrt{x}} - e^{\sqrt{α}}}{\sqrt{x} - \sqrt{α}} \cdot \frac{1}{e^{\sqrt{x}}} \cdot 2$

$= \lim_{x \to α^{+}} \frac{e^{\sqrt{x}} \cdot \frac{1}{2 \sqrt{x}}}{\frac{1}{2 \sqrt{x}}} \cdot \frac{2}{e^{\sqrt{x}}} = 2$

$= \lim_{x \to α^{+}} f (g (x)) = f (\lim_{x \to α^{+}} g (x)) = \sin \frac{π}{6} = \frac{1}{2} = 00.50$

Solution

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Solution

Q. Let α be a positive real number. Let f:ℝ→ℝ and g:(α,∞)→ℝ be the functions defined by f(x)=sin(πx12) and g(x)=2loge(x-α)loge(ex-eα). Then the value of limx→α+f(g(x)) is ___________. [2022]

Ans. (00.50) We have, g(x)=2loge(x-α)loge(ex-eα) Now, limx→α+g(x)=limx→α+2x-α(12x)1ex-eα(12xex), {when x→α+, f(x)→∞∞} Apply L.H. Rule =limx→α+ex-eαx-α·1ex·2 =limx→α+ex·12x12x·2ex=2 =limx→α+f(g(x))=f(limx→α+g(x))=sinπ6=12=00.50

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