Compute the limit

1. Jul 6, 2016

Poznerrr

• Member warned that the homework template is not optional
Hey! Can somebody take a look on these two limit problems? I don't agree with the answer to #15, which is supposed to be 0 while I get infinity. #16 seems to ask to find the value of the sum...I posted a pic of my attempts to solve the problems below.

My attempts:

2. Jul 6, 2016

pasmith

#15 is asking for the limit *from above*. When $\pi/2 < x < \pi$, $\tan x$ is negative.

However you are correct that the limit *from below* is $e^{+\infty} = +\infty$.

3. Jul 6, 2016

Poznerrr

Ooooh, I see! Thanks!

4. Jul 6, 2016

Fightfish

• Poster has been reminded to not do the student's homework for them
For Problem 16, the approach should be to convert the summation into an integral thus:
$$\lim_{n \to \infty} \sum_{i = 1}^{n} \frac{3}{n} \sqrt{9 - \left(\frac{3i}{n}\right)^2} = \lim_{n \to \infty} \sum_{i = 1}^{n} \frac{3 \Delta i}{n} \sqrt{9 - \left(\frac{3i}{n}\right)^2}$$
where $\Delta i = 1$.
Then, it is straightforward to have a variable $\lambda = i / n$, so that in the limit $n \to \infty$, the summation becomes the integral
$$\int_{0}^{1}d\lambda\,\,3\sqrt{9 - 3\lambda^2 }$$

5. Jul 6, 2016

Poznerrr

Thanks! I understand the substitution, but don't quite get how you came up with the limits of integration (0 and 1).

6. Jul 6, 2016

Fightfish

The variable $\lambda = i/n$ takes values between $0$ (i = 1) and $1$ (i = n), which are the corresponding limits of the original summation.

7. Jul 6, 2016

Poznerrr

The limits make sense then. I'm trying to compute the integral and something is really off...

8. Jul 6, 2016

Poznerrr

Here's what I get when I try to compute the integral

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9. Jul 6, 2016

Fightfish

The integration is not correct; you need to use integration by parts - remember its an $u^2$ under the square root.

10. Jul 7, 2016

Poznerrr

I set u=sqrt(1-u^2) (du=-u/sqrt(1-u^2) and dv=du, so that v=u. But this didn't help much, since I still need in integrate sqrt. (1-u^2). It started to look more like inverse sine though.

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• IMG_0236.JPG
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11. Jul 7, 2016

Fightfish

Actually nevermind about the integration by parts; it's much faster to do a substitution directly on the original integral $\int d\lambda \,\,\sqrt{1-\lambda^2}$ - let $\lambda \equiv \sin \theta$ and proceed accordingly.

12. Jul 8, 2016

Poznerrr

Yeah, I used trig sub and got the answer I was supposed to. Thanks for your help!