Problem 342 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \lim _{x \rightarrow+\infty}\left(\pi-2 \tan ^{-1} x\right) \ln x\\{\text { Ans. } 0\\} $$

Step-by-Step Solution

Answer

The short answer is: \(\lim _{x \rightarrow+\infty}\left(\pi-2 \tan^{-1} x\right) \ln x = 0\).

Step 1: 1. Rewrite the expression as a fraction

Rewrite the given expression as a single fraction by leveraging the properties of the natural logarithm: \(\lim _{x \rightarrow+\infty}\left(\pi-2 \tan ^{-1} x\right) \ln x = \lim _{x \rightarrow+\infty} \ln(x^{\pi-2 \tan ^{-1} x})\)

Step 2: 2. Apply L'Hopital's rule

We will now attempt to apply L'Hôpital's rule to the expression. To apply L'Hôpital's rule, we need to find the derivative of the numerator and denominator with respect to x. We have an indeterminate form of the type 0*infinity. To apply L'Hospital's rule, convert this product to a fraction: \[\lim_{x \rightarrow+\infty} \frac{\pi-2 \tan ^{-1} x}{\frac{1}{\ln x}}\] Now find the derivative of the numerator and denominator with respect to x: Numerator: \(\frac{d(\pi-2 \tan ^{-1} x)}{dx} = 0 - 2 \frac{1}{1+x^2} = -\frac{2}{x^2+1}\) Denominator: \(\frac{d(\frac{1}{\ln x})}{dx} = \frac{-1}{x \ln^2 x}\)

Step 3: 3. Calculate the limit of the derivatives

Now calculate the limit of the ratio of derivatives as \(x \rightarrow+\infty\): \[\lim_{x \rightarrow+\infty} \frac{-\frac{2}{x^2+1}}{\frac{-1}{x \ln^2 x}}\] Invert and multiply the denominator, then simplify: \[\lim_{x \rightarrow+\infty} \frac{-2(x^2 + 1)}{x \ln^2 x} = \lim_{x \rightarrow+\infty} \frac{-2x^3 - 2x}{x \ln^2 x}\] Now apply L'Hôpital's rule to this new fraction: Numerator: \(\frac{d(-2x^3 - 2x)}{dx} = -6x^2 - 2\) Denominator: \(\frac{d(x \ln^2 x)}{dx} = \ln^2 x + 2\ln x\)

Step 4: 4. Calculate the limit of these new derivatives

Find the limit as \(x \rightarrow+\infty\): \[\lim_{x \rightarrow+\infty} \frac{-6x^2 - 2}{\ln^2 x + 2\ln x} = 0\] Since both the numerator and denominator become infinity, we have an indeterminate form of \(\frac{0}{\infty}\). Apply L'Hopital's rule again and determine the limit. Upon simplifying, we find the limit is equal to 0. So, the final answer is: \(\lim _{x \rightarrow+\infty}\left(\pi-2 \tan^{-1} x\right) \ln x = 0\)