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How is

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

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

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

$$\begin{align*} \lim_{x\to \infty}x^{x}\sin\left(\frac{b}{x^{x}}\right) &= \lim_{x\to \infty}x^{x}\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)!}\left(\frac{b}{x^{x}}\right)^{2n+1}\\ &=b + \lim_{x\to\infty}\sum_{n=1}^{\infty}\frac{(-1)^{n}}{(2n+1)!}\frac{b^{2n+1}}{x^{2nx}}\\ &=b. \end{align*}$$

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