darknecross
-
Posts
4 -
Joined
-
Last visited
Content Type
Profiles
Forums
Events
Posts posted by darknecross
-
-
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.
0 -
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".
0 -
[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.
0
Funky limit question
in Analysis and Calculus
Posted · Edited by darknecross
typo
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.