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If $\displaystyle \lim_{x\rightarrow 0}\frac{x^n\sin^{n}(x)}{x^{n}-(\sin x)^{n}}$ is a finite non zero number.

Then value of $n$ is equals

What I try:

$$\sin x = x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots $$

$$\lim_{x\rightarrow 0}\frac{x^n\bigg(x-\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots \bigg)^n}{x^{n}-\bigg(x-\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots \bigg)^n}$$

$$\lim_{x\rightarrow 0}\frac{x^{n+1}\bigg(1-\frac{x^2}{3!}+\frac{x^4}{5!}+\cdots \bigg)^n}{x^{n}-\bigg(x-\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots \bigg)^n}$$

How do I solve it? Help me please.

Community
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jacky
  • 5,194

5 Answers5

3

For each $n\in\mathbb N$, the first non-null term of the Taylor series of $x^n\sin^n(x)$ centered at $0$ is $x^{2n}$, whereas the first non-null term of the Taylor series of $x^n-\sin^n(x)$ centered at $0$ is $kx^{n+2}$, for some $k\neq0$. So, your limit will be finite and non zero if and only if $2n=n+2$. And this occurs if and only if $n=2$.

José Carlos Santos
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$$L=\lim_{x \rightarrow 0} \frac{x^n \sin^{n} x}{x^n-\sin^n x}$$ $$= \lim_{x \rightarrow 0} \frac{x^{2n}(1-x^2/6)^n}{x^{n}-x^n(1-x^2/6)^n}$$ $$=\lim_{x \rightarrow 0} \frac{x^n (1-nx^2/6)}{nx^2/6}=\lim_{x \rightarrow 0}\frac {6x^{n-2}}{n}= 3,~ if~ n =2,$$ $$= 0, ~if~ n > 2, $$ $$=\infty ~~if~~n= n <2 $$

Z Ahmed
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The denominator factors as

$$(x-\sin x)\sum_{k=0}^{n-1} x^{n-k-1}\sin^kx\sim ax^3x^{n-1}$$ while the numerator is $\sim bx^{2n}$.

So $2n=n+2$.

1

$$\frac{\sin x}{x}\simeq 1-\frac{1}{6}x^2+\frac{1}{120}x^4+O(x^6)$$ $$\left(\frac{\sin x}{x}\right)^n\simeq 1-\frac{n}{6}x^2+\frac{n(5n-2)}{360}x^4+O(x^6)$$ $$\frac{\left(\frac{\sin x}{x}\right)^n}{1-\left(\frac{\sin x}{x}\right)^n}\simeq \frac{6}{n}x^{-2}-\frac{5n+2}{10n}+O(x^2)$$ $$\boxed{f(x)=\frac{x^n\sin^n x}{x^n-sin^nx}\simeq\frac{6}{n}x^{n-2}-\frac{5n+2}{10n}x^n+O(x^{n+2})}$$

If $\quad n<2\qquad$ $f(x)\sim\frac{6}{n}\frac{1}{x^{2-n}}\to\infty$

If $\quad n=2\qquad f(x)\sim 3-\frac35 x^2\to 3$

If $\quad n>2\qquad f(x)\sim\frac{6}{n}x^{n-2}\to 0 $

JJacquelin
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Assuming $n\neq 0$ (to make the expression under limit meaningful) the given expression can be written as $$\dfrac{\dfrac{x} {\sin x} - 1}{\left(\dfrac{x}{\sin x}\right)^n-1}\cdot\frac{x^3}{x-\sin x} \cdot x^{n-2}$$ The first factor tend to $1/n$ and second one tends to $6$ and hence the limiting behavior of the expression crucially depends on that of $x^{n-2}$. The desired limit is thus $6/n=3$ if $n=2$, is $0$ if $n>2$ and does not exist if $n<2$.

Paramanand Singh
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