/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 35 If \(\lim _{x \rightarrow 0} \fr... [FREE SOLUTION] | 91Ó°ÊÓ

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If \(\lim _{x \rightarrow 0} \frac{x^{2 n} \sin ^{n} x}{x^{2 n}-\sin ^{2 n} x}\) is a non zero finite number, then n must be equal to (A) 1 (B) 2 (C) 3 (D) none of these

Short Answer

Expert verified
Answer: (D) none of these

Step by step solution

01

Apply L'Hopital's Rule

We need to differentiate the numerator and the denominator with respect to x. So, $$ \frac{d}{d x}\left(x^{2 n} \sin ^{n} x\right) = 2nx^{2n-1}\sin^n{x} + nx^{2n}\sin^{n-1}{x} \cos{x} $$ and $$ \frac{d}{d x}\left(x^{2 n}-\sin ^{2 n} x\right) = 2nx^{2n-1} - 2n\sin^{2n-1}{x} \cos{x} $$ Since we need the limit as x approaches 0, let's put x = 0 in the derivatives: $$ \lim _{x \rightarrow 0} \frac{d\left(x^{2 n} \sin ^{n} x\right)/dx}{d\left(x^{2 n}-\sin ^{2 n} x\right)/dx} = \lim _{x \rightarrow 0} \frac{ 2nx^{2n-1}\sin^n{x} + nx^{2n}\sin^{n-1}{x} \cos{x}}{2nx^{2n-1} - 2n\sin^{2n-1}{x} \cos{x}} $$
02

Simplify the Limit Expression

To simplify the limit, let's factor out the common terms: $$ \lim _{x \rightarrow 0} \frac{n(x^{2n}(1+\sin^{2n-1}{x}\cos{x})}{2n(x^{2n-1}(1-\sin^{2n-1}{x}\cos{x})} $$ Now, cancel the common terms of n and x^(2n-1) to get: $$ \lim _{x \rightarrow 0} \frac{1 + \sin^{2n-1}{x}\cos{x}}{2(1 - \sin^{2n-1}{x}\cos{x})} = \frac{1}{2} $$ Since the limit is a non-zero finite number, our analysis shows that: n must be equal to (D) none of these.

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Most popular questions from this chapter

Let \(l_{1}=\lim _{x \rightarrow \infty} \sqrt{\frac{x-\cos ^{2} x}{x+\sin x}}\) and $l_{2}=\lim _{\mathrm{h} \rightarrow 0^{-}} \int_{-1}^{1} \frac{\mathrm{h} \mathrm{dx}}{\mathrm{h}^{2}+\mathrm{x}^{2}} .$ Then (A) both \(l_{1}\) and \(l_{2}\) are less than \(\frac{22}{7}\) (B) one of the two limits is rational and other irrational. (C) \(l_{2}>l_{1}\) (D) \(l_{2}\) is greater than 3 times of \(l_{1}\).

Which of the following limits exist? (where [.] indicates greatest integer function all throughout) (A) \(\lim _{x \rightarrow 1} \frac{\sin [x]}{[x]}\) (B) $\lim _{n \rightarrow \infty}\left(\frac{\mathrm{e}^{\mathrm{n}}}{\pi}\right)^{1 / \mathrm{n}}$ (C) \(\lim _{x \rightarrow 1}\left[\sin \left(\sin ^{-1} x\right)\right]\) (D) \(\lim _{x \rightarrow \pi / 2}\left[\sin ^{-1}(\sin x)\right]\)

Column - I (A) \(\lim _{x \rightarrow 0} \frac{1-\cos 2 x}{e^{x^{3}}-e^{x}+x}\) equals (B) If the value of $\lim _{\mathrm{x} \rightarrow 0^{-}}\left(\frac{(3 / \mathrm{x})+1}{(3 / \mathrm{x})-1}\right)^{1 / \mathrm{x}}$ can be expressed in the form of \(\mathrm{e}^{\mathrm{iq}}\), where \(\mathrm{p}\) and \(\mathrm{q}\) are relative prime then \((\mathrm{p}+\mathrm{q})\) is equal to (C) \(\lim _{x \rightarrow 0} \frac{\tan ^{3} x-\tan x^{3}}{x^{5}}\) equals (D) $\lim _{x \rightarrow 0} \frac{x+2 \sin x}{\sqrt{x^{2}+2 \sin x+1}-\sqrt{\sin ^{2} x-x+1}}$ Column-II (P) 1 (Q) 2 (R) 4 (S) 5

Assertion \((\mathbf{A}):\) Let \(\mathrm{f}:(0, \infty) \rightarrow \mathrm{R}\) be a twice continuously differentiable function such that $\left|f^{\prime}(x)+2 x f^{\prime}(x)+\left(x^{2}+1\right) f(x)\right| \leq 1\( for all \)x$ Then \(\lim _{x \rightarrow \infty} f(x)=0\). Reason (R) : Applying L'Hospital's rule twice on the function $\frac{f(x) e^{\frac{x^{2}}{2}}}{e^{\frac{x^{3}}{2}}}\( we get \)\lim _{x \rightarrow \infty} f(x)=0$.

If $\lim _{x \rightarrow \infty}\left(\sqrt{a^{2} x^{2}+a x+1}-\sqrt{a^{2} x^{2}+1}\right)$ $=\mathrm{K} \cdot \lim _{x \rightarrow \infty}(\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x})\( then the value of \)\mathrm{K}$ (A) (B) (C) \(2 \mathrm{a}\) (D) None of these

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