Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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

share|improve this question
1  
Sorry, I removed the $a$ by mistake. –  Gigili Jul 29 '12 at 16:05

5 Answers 5

up vote 5 down vote accepted

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 $$ \lim_{t\rightarrow\infty}\left(\left(1+\frac{1}{t}\right)^{t}\right)^a=e^a\\ $$

share|improve this answer

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

share|improve this answer
    
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 –  Bunuelian Trick Jul 29 '12 at 16:08
    
@i.m.soloveichik true I hope.. –  Bunuelian Trick Jul 29 '12 at 16:09
4  
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.

share|improve this answer

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 $e$.

share|improve this answer

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 $4$.

$\implies a=\log_e4$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.