# Evaluate the limit of the given problem

• chwala
chwala
Gold Member
Homework Statement
Evaluate ##\lim_{x→0^+} (\sin x)^x##

(Textbook example).
Relevant Equations
Limits

Ok, to my question; was the step highlighted necessary? The working steps to solution are clear. Cheers.

The line before is
$$\lim_{x \rightarrow 0^+} \frac{-x^2}{\tan x}$$
which is the indeterminate form 0/0, so L'Hôpital's rule has to be applied again.

chwala
Alternatively $$(\sin x)^x = \left(\frac{\sin x}x\right)^x x^x = \exp\left(x\ln\left(\frac{\sin x}x\right) + x \ln x\right)$$ and use that $$\begin{split} \lim_{x \to 0} \frac{ \sin x}{x} &= \sin'(0) = 1 \\ \lim_{x \to 0^{+}} x \ln x &= 0 \end{split}$$

chwala
Okay...I missed that.... We are avoiding to divide by ##0## on denominator that will give (indeterminate form).

Thus
##\lim_{x→0^+} [-2x\cos^2 x] =0##

Cheers.

Last edited:
chwala said:
Okay...I missed that.... We are avoiding to divide by ##0## on denominator that will give (indeterminate form). Thus
##-2x\cos^2 x## on numerator.
Cheers.
It's not just that the denominator is zero in the limit -- it's because the numerator is also approaching zero. Thus it's the ##[\frac 0 0]## indeterminate form, as @DrClaude mentioned.

chwala and docnet
Mark44 said:
It's not just that the denominator is zero in the limit -- it's because the numerator is also approaching zero. Thus it's the ##[\frac 0 0]## indeterminate form, as @DrClaude mentioned.
@Mark44 hi... I understand that too. Thks mate... My problem was on the highlighted part in red. Taking limits as indicated for my expression in post ##4## will realise the given value ##0##.

Gavran
chwala said:
@Mark44 hi... I understand that too. Thks mate... My problem was on the highlighted part in red. Taking limits as indicated for my expression in post ##4## will realise the given value ##0##.
There are only two cases where L’Hospital’s rule can be applied, ## \frac{0}{0} ## and ## \frac{\infty}{\infty} ##, and both of them you have in your example.

chwala
pasmiths argument seems possibly to justify my feeling that since sin(x) is tangent at 0 to x, the limit should be the same as that of x^x? (i.e. 1). maybe too optimistic/lazy.

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