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For $x>0$, $$\lim_{x \rightarrow 0} \left[ \sin(x)^{\frac{1}{x}}+\left(\frac{1}{x}\right)^{\sin(x)}\right] $$These are two forms of $0^\infty$ and $\infty^0$. I know these are to be evaluated separately and then added. But how do I start?

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Is your question solved in a satisfying way? If no tell us what's wrong, if yes, could you be so kind to accept an answer? :) –  Dominic Michaelis Feb 17 '13 at 12:02
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Sorry, for being late to respond... @Dominic: Yes, I am satisfied. Thanks a lot! –  Ashish Gaurav Feb 17 '13 at 13:56

2 Answers 2

up vote 4 down vote accepted

At first start with $$a^b=e^{b\cdot \ln(a)}$$ (this is the definition of $a^b$ for $a>0$).
and because of the $e$ function is continuous $$\lim e^{x_n}=e^{\lim x_n}$$ This is because of the sequence definition or test for continuous functions, a function $f$ is continuous if for every convergent sequence $x_n$ $$ \lim_{n\rightarrow \infty} f(x_n)=f(\lim_{n\rightarrow \infty} x_n)$$ Since both term are greater equal we can check the first, and afterwards the second. $$\sin(x)^\frac{1}{x}=e^{\ln(\sin(x))\cdot \frac{1}{x}}$$ Because of it is continuous we check $$\lim_{x\rightarrow 0}\frac{\ln(\sin(x))}{x}$$ As we have a $\frac{-\infty}{\infty}$ expression here, we can use L'hospital $$\lim_{x\rightarrow 0 } \frac{\ln(\sin(x))}{x}=-\lim_{x\rightarrow 0} \frac{\cos(x)}{\sin(x)}=-\infty$$ (the $-$ must be there because $\ln(\sin(x))<0$ as $x\rightarrow 0$
And because $$\lim_{x\rightarrow - \infty} e^{x}=0$$ the first term vanishes.

The second Term is handled like this: $$\left(\frac{1}{x}\right)^{\sin(x)} = e^{\ln\left(\frac{1}{x}\right) \sin(x)}$$

Now $$\ln\left(\frac{1}{x}\right) \cdot \sin(x)=\frac{\ln\left(\frac{1}{x}\right)}{\frac{1}{\sin(x)}}$$ with L'hospital $$\lim_{x\rightarrow 0} \frac{\ln\left(\frac{1}{x}\right)}{\frac{1}{\sin(x)}}= \lim_{x\rightarrow 0} \frac{x}{\cos(x)}=0 $$ so the second term is $e^0$ so the limit is 1.

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could you link to a proof of that, didn't know that before. –  yiyi Feb 17 '13 at 10:31
    
of which exactly ? –  Dominic Michaelis Feb 17 '13 at 10:33
    
That because $e$ is a continuous function that $\lim e^{x_n}=e^{\lim x_n}$ I didn't know that if $f\left(h\left(x\right)\right)$ is continuous then $\lim f\left(h\left(x\right)\right)$ = $f\left(\lim h\left(x\right)\right)$ –  yiyi Feb 17 '13 at 10:37
    
ok i added the stuff, as i learned it in university this was the definition of continuous, which one is yours, so i can link the right proof ? –  Dominic Michaelis Feb 17 '13 at 10:40
    
I am not at university yet. –  yiyi Feb 17 '13 at 10:49

Note that $0^{\infty}$ is not an indeterminate form. Then $$\lim_{x \rightarrow 0^+} \left[ \sin(x)^{\frac{1}{x}}+\left(\frac{1}{x}\right)^{\sin(x)}\right]=\lim_{x \rightarrow 0^+} \left(\frac{1}{x}\right)^{\sin(x)}=\lim_{y \rightarrow +\infty}y^{\sin(1/y)}=\lim_{y \rightarrow +\infty}(y^{1/y})^{y\sin(1/y)}=1$$ The other limit to $0^-$is complex infinity.

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is $0^\infty$ really not an indeterminate form? I mean $\lim_{n \rightarrow \infty} 0^n $ is obviously 0 –  Dominic Michaelis Feb 17 '13 at 11:28
    
@DominicMichaelis: I didn't learn at school that this is an indeterminate case. Or is it? Things should be clear (btw). –  Chris's sis Feb 17 '13 at 11:34
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It's indeterminate, $$\lim_{x\rightarrow 0} e^{{-x^{-2}}^\frac{1}{x}}$$ should be $0$ –  Dominic Michaelis Feb 17 '13 at 11:38
    
@DominicMichaelis: can you show me this in any calculus book? –  Chris's sis Feb 17 '13 at 11:39
    
Are you serious? I gave you a counterexample for $0^\infty$ is determinate, in the limit above it is $1$ in my example it is $0$, shall i prove that the limit of my example is $0$ ? –  Dominic Michaelis Feb 17 '13 at 11:41

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