Evaluate the limit of the given problem

[chwala]

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.

 

[DrClaude]

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.

 

[pasmith]

Alternatively <br /> (\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}<br /> \lim_{x \to 0} \frac{ \sin x}{x} &amp;= \sin&#039;(0) = 1 \\<br /> \lim_{x \to 0^{+}} x \ln x &amp;= 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.

 

[Mark44]

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]

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]

@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.

 

[mathwonk]

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.