Prove if $$\lim_{x\to\pi/2}\frac{\sin x-(\sin x)^{\sin x}}{1-\sin x+\log_e \sin x}=2$$ is true or not.
Here we have L.H.S. in $\frac{0}{0}$ form so I can use L'Hospital's rule. But applying it only leads to complicated and messier terms.
Series expansion also does not seem to help for me as I don't know the expansion of $\sin x^{\sin x}$ or of $\log_e \sin x$.
Is there any other method to evalute this limit?
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1In the numerator, do you mean $(\sin x)^{\sin x}$ or $\sin (x^{\sin x})$ ? – JSCB Mar 12 '13 at 13:40
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2Substitute $t$ for $\sin(x)$. Then try L'Hopital. – Macavity Mar 12 '13 at 13:42
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is a substitution allowed? $\sin(x)$ is not injective in any neighbourhood of $\frac{\pi}{2}$ My first idea would be to write it in cos than you have the limit to 0 which makes the series easier – Dominic Michaelis Mar 12 '13 at 13:45
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Yes, substitution is allowed. – Berci Mar 12 '13 at 14:21
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A related problem. – Mhenni Benghorbal Mar 12 '13 at 14:23
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First, substitute $t=x-\pi/2$. Then you get $\sin x=\cos t$. Your limit turns into: $$\lim_{t\to0}\frac{\cos t-(\cos t)^{\cos t}}{1-\cos t+\ln \cos t}$$ Now, the easiest would be to use Taylor polynomials around $0$: $$\cos t=1+t^2/2+o(t^2), \hspace{8pt} \ln(1+x)=x+x^2/2+o(x^2), \hspace{8pt} e^x=1+x+x^2/2+o(x^2)$$ Also, remember that $(\cos t)^{\cos t}=e^{\cos t\ln\cos t}$
Dennis Gulko
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