# How is $\underset{x\rightarrow{a}}{\lim}a^x\sin(b/a^x)=b$? [closed]

How is

$$\underset{x\to a}{\lim}a^x\sin\bigg(\frac b{a^x}\bigg)=b$$

## closed as off-topic by Paul Frost, callculus, Adrian Keister, Lee David Chung Lin, Paramanand SinghApr 15 at 18:52

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• It isn't. @hammad – Saucy O'Path Apr 15 at 12:58
• Put x=a in the expression – Tojrah Apr 15 at 13:01
• Are you certain that the problem says $\lim_{x\to a}$, and not, say, $\lim_{x\to\infty}$? – Arthur Apr 15 at 13:11
• @HammadAhmed Judging from the options, the limit process should be $x→+∞$ if $a>1$. – Saad Apr 15 at 13:11
• @HammadAhmed Since $f(x):=a^x\sin\left(\dfrac b{a^x}\right)$ is continuous at $x=a$, then $\lim\limits_{x→a}f(x)=f(a)$. – Saad Apr 15 at 13:22

There is probably a typo here. Suppose that for some choice of $$a,b$$ with $$a\neq 0$$ we have $$\lim_{x\to a}a^{x}\sin\left(\frac{b}{a^{x}}\right) = b.$$ Then by continuity this means that $$a^{a}\sin\left(\frac{b}{a^{a}}\right) = b,$$ which implies that $$\sin\left(\frac{b}{a^{a}}\right) = \frac{b}{a^{a}}.$$ However, the only place where $$\sin(x) = x$$ is at $$x = 0$$, so it must be the case that $$b = 0$$ which in some sense trivializes the problem. I suspect that the limit should actually read
$$\lim_{x\to \infty}x^{x}\sin\left(\frac{b}{x^{x}}\right) = b.$$
This we can prove using the Taylor series expansion for $$\sin(x)$$:
\begin{align*} \lim_{x\to \infty}x^{x}\sin\left(\frac{b}{x^{x}}\right) &= \lim_{x\to \infty}x^{x}\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)!}\left(\frac{b}{x^{x}}\right)^{2n+1}\\ &=b + \lim_{x\to\infty}\sum_{n=1}^{\infty}\frac{(-1)^{n}}{(2n+1)!}\frac{b^{2n+1}}{x^{2nx}}\\ &=b. \end{align*}