JEE Challenger
More from Limits and Derivatives
MathMedAdv 2022 • Paper 1

Limit of Composite Function involving Logarithmic and Trigonometric Terms

Let α\alpha be a positive real number. Let f:R→Rf: \mathbb{R} \rightarrow \mathbb{R} and g:(α,∞)→Rg: (\alpha, \infty) \rightarrow \mathbb{R} be the functions defined by

f(x)=sin⁡(πx12)andg(x)=2log⁡e(x−α)log⁡e(ex−eα).f(x) = \sin\left(\frac{\pi x}{12}\right) \quad \text{and} \quad g(x) = \frac{2 \log_e (\sqrt{x} - \sqrt{\alpha})}{\log_e (e^{\sqrt{x}} - e^{\sqrt{\alpha}})}.

Then the value of lim⁡x→α+f(g(x))\lim_{x \rightarrow \alpha^+} f(g(x)) is __________ .

Official Numerical Answer0.5

Topics & Concepts

Limits and DerivativesLimits

Step-by-Step Solution

To find the value of the limit lim⁡x→α+f(g(x))\lim_{x \rightarrow \alpha^+} f(g(x)), we first determine the limit of g(x)g(x) as x→α+x \rightarrow \alpha^+.

The given function g(x)g(x) is defined as: g(x)=2log⁡e(x−α)log⁡e(ex−eα)g(x) = \frac{2 \log_e (\sqrt{x} - \sqrt{\alpha})}{\log_e (e^{\sqrt{x}} - e^{\sqrt{\alpha}})}

Let t=x−αt = \sqrt{x} - \sqrt{\alpha}. As x→α+x \rightarrow \alpha^+, we have t→0+t \rightarrow 0^+. Thus, x=α+t\sqrt{x} = \sqrt{\alpha} + t.

Substituting tt into the denominator of g(x)g(x): log⁡e(ex−eα)=log⁡e(eα+t−eα)=log⁡e(eα(et−1))\log_e (e^{\sqrt{x}} - e^{\sqrt{\alpha}}) = \log_e \left( e^{\sqrt{\alpha} + t} - e^{\sqrt{\alpha}} \right) = \log_e \left( e^{\sqrt{\alpha}} (e^t - 1) \right)

Using the logarithmic identity log⁡e(ab)=log⁡ea+log⁡eb\log_e (ab) = \log_e a + \log_e b: log⁡e(ex−eα)=log⁡e(eα)+log⁡e(et−1)=α+log⁡e(et−1)\log_e (e^{\sqrt{x}} - e^{\sqrt{\alpha}}) = \log_e (e^{\sqrt{\alpha}}) + \log_e (e^t - 1) = \sqrt{\alpha} + \log_e (e^t - 1)

Now, we can express et−1e^t - 1 as t⋅(et−1t)t \cdot \left(\frac{e^t - 1}{t}\right), which yields: log⁡e(et−1)=log⁡et+log⁡e(et−1t)\log_e (e^t - 1) = \log_e t + \log_e \left(\frac{e^t - 1}{t}\right)

Substituting this back into the expression for g(x)g(x): g(x)=2log⁡etα+log⁡et+log⁡e(et−1t)g(x) = \frac{2 \log_e t}{\sqrt{\alpha} + \log_e t + \log_e \left(\frac{e^t - 1}{t}\right)}

Dividing the numerator and the denominator by log⁡et\log_e t (since t→0+t \rightarrow 0^+, log⁡et→−∞≠0\log_e t \rightarrow -\infty \neq 0): g(x)=21+αlog⁡et+log⁡e(et−1t)log⁡etg(x) = \frac{2}{1 + \frac{\sqrt{\alpha}}{\log_e t} + \frac{\log_e \left(\frac{e^t - 1}{t}\right)}{\log_e t}}

Now, taking the limit as t→0+t \rightarrow 0^+:

  1. lim⁡t→0+αlog⁡et=0\lim_{t \rightarrow 0^+} \frac{\sqrt{\alpha}}{\log_e t} = 0
  2. Since lim⁡t→0+et−1t=1\lim_{t \rightarrow 0^+} \frac{e^t - 1}{t} = 1, we have lim⁡t→0+log⁡e(et−1t)=log⁡e1=0\lim_{t \rightarrow 0^+} \log_e \left(\frac{e^t - 1}{t}\right) = \log_e 1 = 0. Therefore, lim⁡t→0+log⁡e(et−1t)log⁡et=0−∞=0\lim_{t \rightarrow 0^+} \frac{\log_e \left(\frac{e^t - 1}{t}\right)}{\log_e t} = \frac{0}{-\infty} = 0.

Thus, we find: lim⁡x→α+g(x)=21+0+0=2\lim_{x \rightarrow \alpha^+} g(x) = \frac{2}{1 + 0 + 0} = 2

Since f(x)=sin⁡(πx12)f(x) = \sin\left(\frac{\pi x}{12}\right) is a continuous function on R\mathbb{R}: lim⁡x→α+f(g(x))=f(lim⁡x→α+g(x))=f(2)\lim_{x \rightarrow \alpha^+} f(g(x)) = f\left(\lim_{x \rightarrow \alpha^+} g(x)\right) = f(2)

Substituting x=2x = 2 into f(x)f(x): f(2)=sin⁡(π⋅212)=sin⁡(π6)=12=0.5f(2) = \sin\left(\frac{\pi \cdot 2}{12}\right) = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} = 0.5

The value of lim⁡x→α+f(g(x))\lim_{x \rightarrow \alpha^+} f(g(x)) is 0.50.5.

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