# Need Help with limits questions

• Sep 7th 2008, 11:31 AM
nic42991
Need Help with limits questions

1. $\displaystyle lim((1/(2+x)-(1/2))/x)$as x approaches 0

2. $\displaystyle lim(sin(2x)/x)$ as x appproaches 0

I apologize if it is hard to read, but how do you make it easier to read? Is there a program that helps you do that, or do you just use the math key so much that you got used to writing in that "language."
• Sep 7th 2008, 11:37 AM
Jhevon
Quote:

Originally Posted by nic42991

1. $\displaystyle lim((1/(2+x)-(1/2))/x)$as x approaches 0

see here

Quote:

2. $\displaystyle lim(sin(2x)/x)$ as x appproaches 0
hint: multiply by 2/2, we get $\displaystyle 2 \lim_{x \to 0} \frac {\sin (2x)}{(2x)}$

now what?

Quote:

I apologize if it is hard to read, but how do you make it easier to read? Is there a program that helps you do that, or do you just use the math key so much that you got used to writing in that "language."
see here
• Sep 7th 2008, 11:44 AM
Matt Westwood
Quote:

Originally Posted by nic42991
I apologize if it is hard to read, but how do you make it easier to read? Is there a program that helps you do that, or do you just use the math key so much that you got used to writing in that "language."

You get used to writing in it, if you do a lot of it. It's just another life skill to have and put on a CV: fluency in LaTeX, like any other programming or typesetting language, or html or whatever. I taught myself by writing everything I knew about mathematics in LaTeX. (And three sheets of paper later, after lunch I went and did something else.)
• Sep 7th 2008, 12:07 PM
nic42991
Quote:

Originally Posted by Jhevon

hint: multiply by 2/2, we get $\displaystyle 2 \lim_{x \to 0} \frac {\sin (2x)}{(2x)}$

now what?

I know that $\displaystyle \lim_{x \to 0} \frac {\sin (x)}{(x)}$ is 1.
Therefore, $\displaystyle \lim_{x \to 0} \frac {\sin (2x)}{(2x)}$ must be 1 as well, right?
Then you multiply the limit by 2, so the ans. is 2.

Did I prove it correctly?
• Sep 7th 2008, 12:13 PM
Jhevon
Quote:

Originally Posted by nic42991
I know that $\displaystyle \lim_{x \to 0} \frac {\sin (x)}{(x)}$ is 1.
Therefore, $\displaystyle \lim_{x \to 0} \frac {\sin (2x)}{(2x)}$ must be 1 as well, right?
Then you multiply the limit by 2, so the ans. is 2.

Did I prove it correctly?

yes.

of course, sound more "mathy" when you say it :D

also, if the 2x bothers you, you can replace it. let t = 2x say. then $\displaystyle t \to 0$ as $\displaystyle 2x \to 0$, and your limit becomes

$\displaystyle 2 \lim_{t \to 0} \frac {\sin t}t$

which looks a lot more like the limit you know. this is just cosmetics though, not necessary. you are more than welcome to say $\displaystyle \lim_{x \to 0} \frac {\sin (2x)}{(2x)} = 1$