# Finding $a$ from $\lim\limits_{x\rightarrow0}(1+a\sin x)^{\csc x} =4$

The question is to find the value of $a$ from the following equation:

$$\lim_{x\rightarrow 0}(1+a\sin x)^{\csc x} =4$$

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Sorry, I removed the $a$ by mistake. – Gigili Jul 29 '12 at 16:05

## 5 Answers

You have $$\lim_{x\rightarrow 0}\left(1+a\sin x\right)^\frac{1}{\sin x}=\\ \lim_{x\rightarrow 0}\left(\left(1+a\sin x\right)^\frac{1}{a\sin x}\right)^a$$ if you set $t=1/(a\sin x)$, and given that $t$ tends to $\infty$ when $x$ tends to $0$, you have $$\lim_{t\rightarrow\infty}\left(\left(1+\frac{1}{t}\right)^{t}\right)^a=e^a\\$$

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Hint for your problem: $\lim_{n\rightarrow 0}(1+n)^{1/n}=e$

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Using this hint you get $a=2ln(2)$ – i. m. soloveichik Jul 29 '12 at 16:06
Thanks, solved it :) – HackToHell Jul 29 '12 at 16:08
@HackToHell okay – La Belle Noiseuse Jul 29 '12 at 16:08
@i.m.soloveichik true I hope.. – La Belle Noiseuse Jul 29 '12 at 16:09
That will work, but a more generally useful technique would be to take the logarithm. However, I give you (and only you) +1 for just providing a hint and not a complete answer. – Harald Hanche-Olsen Jul 29 '12 at 16:10

Another possible way to do it: take the logarithm of both sides. Since it's a continuous function, you can put it inside the limit and use the logarithm properties to take $\csc x$ outside and then use L'Hôpital's:

$$\lim_{x \to 0} (1+a \sin x)^{\frac1{\sin x}} = 4$$

$$\lim_{x \to 0} \ln [(1+a \sin x)^{\frac1{\sin x}} ] = \ln 4$$

$$\lim_{x \to 0} \frac{\ln(1+a\sin x)}{\sin x} = \ln 4$$

This last limit is a $0/0$ indetermination, and L'Hôpital's solves it easily.

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Let $u = a \sin x$. Then $$(1+a \sin x)^{\csc x} = (1 + u)^{\frac{a}{u}} = ((1+u)^{\frac{1}{u}})^a$$ and we note that as $x \to 0$ we have $u \to 0$. Then we want to solve $$4 = \lim_{u \to 0} ((1+u)^{\frac{1}{u}})^a = (\lim_{u \to 0}(1+u)^{\frac{1}{u}})^a = e^a.$$

Taking logarithms of both sides we find $a = \log 4$, where the logarithm is taken base $e$.

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Let A = $(1+a\sin x)^{\csc x}$

$\log A=\csc x \cdot\log(1+a\sin x)=\dfrac{\log(1+a\sin x)}{\sin x}$

$\lim_{x\rightarrow 0}\log A=a\lim_{x\rightarrow 0}\dfrac{\log(1+a\sin x)}{a\sin x}=a$ as $a\sin x \rightarrow 0\ as\ x\rightarrow 0$

$\implies A=e^a$ which is equal to $4$.

$\implies a=\log_e4$

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