Problem 342 $$ \lim _{x \rightarrow+\infty... [FREE SOLUTION]

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Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Sanjiva Dayal IIT Kanpur

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Chapter 2: Problem 342

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

Short Answer

Expert verified

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

Step by step solution

1. Rewrite the expression as a fraction

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

2. Apply L'Hopital's rule

We will now attempt to apply L'H么pital's rule to the expression. To apply L'H么pital'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: $\frac{d(\frac{1}{\ln x})}{dx} = \frac{-1}{x \ln^2 x}$

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'H么pital's rule to this new fraction: Numerator: $\frac{d(-2x^3 - 2x)}{dx} = -6x^2 - 2$ Denominator: $\frac{d(x \ln^2 x)}{dx} = \ln^2 x + 2\ln x$

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

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91影视

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

Short Answer

Step by step solution

1. Rewrite the expression as a fraction

2. Apply L'Hopital's rule

3. Calculate the limit of the derivatives

4. Calculate the limit of these new derivatives

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