darknecross

Say we have [math]\lim_{x\to a} n[/math] to make this incredibly vague; If [math]\lim_{x\to a^{+}} n \neq \lim_{x\to a^{-}} n[/math] then we say the limit doesn't exist at n. Which makes sense, since either side of the graph is approaching a different value. If you're trying to find the right/left-hand limit, then that's fine. But if it's not specified, it's assumed to be from both sides. For this problem, I'd say that [math]\lim_{x\to 0^{+}}\; x^{\sin x} \; = \; 1[/math] The domain has nothing to do with the limit laws, if that was implied. I'm guessing the problem was looking for the answer of 1 regardless, and the right-hand limit wasn't important in evaluation.

Okay, I'm seeing it now. The limit laws only work when [math]\lim_{x\to a}f(x)[/math] and [math]\lim_{x\to a}g(x)[/math] both exist. In this case, [math]\lim_{x\to 0}\; x\ln^{2} x[/math] doesn't exist, so you can't use that limit law.

Yeah, you can distribute the limit, but you can't take them individually. For the mathematical reason, I'm not entirely sure of the details, other than my Calc 1 professor used to say "I could prove a lot of untrue things if I did that".

[math] \lim_{x\to 0} x^{\sin x} [/math] [math]0^{0}[/math] is indeed indeterminate, so yeah, rewrite it. [math] \lim_{x \to 0}\;e^{\displaystyle \ln x^{\sin x}} [/math] You use the ln to bring the sinx out in front, but you need the e to balance that. A lot of people just take the natural log of both sides, which works too. In my notes/homework I like working across the page instead of having a new line, and this reduces clutter. [math] e^{\displaystyle \lim_{x\to0}\;\sin x \ln x} [/math] In this step, I bring the sinx out before the ln, and the limit inside the exponent. Notice you have [math]0\times-\infty[/math] [math] e^{\displaystyle \lim_{x\to0}\;\frac{\ln x}{\frac{1}{\sin x}}} [/math] We're dealing with the [math]\frac{\infty}{\infty}[/math] form, so we use l'Hopital's rule. [math] e^{\displaystyle \lim_{x\to0}\;\frac{\frac{1}{x}}{\frac{-\cos x}{sin^{2}x}}} [/math] Simplified... [math] e^{\displaystyle \lim_{x\to0}\;\frac{sin^{2}x}{-x\cos x}} [/math] Now we have [math]\frac{0}{0}[/math] so we use l'Hopital's rule again. [math] e^{\displaystyle \lim_{x\to0}\;\frac{2\sin x \cos x}{x\sin x - \cos x}} [/math] The limit approaches [math]\frac{0}{-1}[/math], or just 0 So the answer is... [math] e^{0}\;\;=\;\; 1 [/math] It's also important to note that for the function [math]f(x)=x^{\sin x}[/math] the Domain is [math](0,\infty)[/math], so technically [math]\lim_{x\to0}\;x^{\sin x}[/math] doesn't exist, only [math]\lim_{x\to0^{+}}[/math] does. Cap'n Refsmmat, I found a big mistake in your answer. [math]-1 \times \lim_{x\to 0} \cos x \times \lim_{x\to 0} (\ln x)^2 \times x \neq \; -1 \times \lim_{x\to 0} (\ln x)^2 \times x [/math] You cannot take the limit of part of the equation, you must wait until the end to take the entire limit at one time.