# Limits and Continuity

• Oct 1st 2008, 10:44 PM
macgyver
Limits and Continuity

x approaching zero.

I'm supposed to determine the limit graphically and then confirm algebraically.

Thank you.
• Oct 1st 2008, 10:49 PM
Chris L T521
Quote:

Originally Posted by macgyver

x approaching zero.

I'm supposed to determine the limit graphically and then confirm algebraically.

Thank you.

Here's the graph:

http://img.photobucket.com/albums/v4...T521/limit.jpg

Can you determine $\displaystyle \lim_{x\to 0}\frac{\sin(2x)}{x}$ now?

Here's a hint on how to verify it algebraically.

$\displaystyle \lim_{x\to 0}\frac{\sin(2x)}{x}=\lim_{x\to 0}\frac{2\sin(2x)}{2x}=2\lim_{x\to 0}\frac{\sin(2x)}{2x}$

This limit should look pretty familiar...

Does this help?

--Chris
• Oct 1st 2008, 10:50 PM
wisterville
Hello,

I don't know what you mean by "graphically"(draw a graph? computing areas?).
For the "algebraically" one, do you know lim((sin x)/x)=1?
If so, transform as (sin 2x)/x=2(sin 2x)/(2x) and let 2x go to 0.

Bye.

P.S. Seems I was slower than Chris.
• Oct 1st 2008, 11:06 PM
macgyver
Quote:

Originally Posted by Chris L T521
Here's the graph:

http://img.photobucket.com/albums/v4...T521/limit.jpg

Can you determine $\displaystyle \lim_{x\to 0}\frac{\sin(2x)}{x}$ now?

Here's a hint on how to verify it algebraically.

$\displaystyle \lim_{x\to 0}\frac{\sin(2x)}{x}=\lim_{x\to 0}\frac{2\sin(2x)}{2x}=2\lim_{x\to 0}\frac{\sin(2x)}{2x}$

This limit should look pretty familiar...

Does this help?

--Chris

Thank you very much for responding Chris. I only get part of it though. could you please tell me how to determine the limit just by looking at the graph? and also how do I solve the rest of the equation. I am very new to calculus so I would really appreciate your help.
• Oct 1st 2008, 11:11 PM
Chris L T521
Quote:

Originally Posted by macgyver
Thank you very much for responding Chris. I only get part of it though. could you please tell me how to determine the limit just by looking at the graph? and also how do I solve the rest of the equation. I am very new to calculus so I would really appreciate your help.

To find the limit graphically, look at the value of the graph at $\displaystyle x=0$. In this case it appears to be 2, although it doesn't exist at 0 [otherwise we would be dividing by zero, which is illegal].

To finish it off, this is what we would do algebraically:

$\displaystyle \lim_{x\to 0}\frac{\sin(2x)}{x}=\lim_{x\to 0}\frac{2\sin(2x)}{2x}=2\lim_{x\to 0}\frac{\sin(2x)}{2x}$

This limit has the form of $\displaystyle \lim_{u\to 0}\frac{\sin(u)}{u}=1$

So let $\displaystyle u=2x$. Thus, as $\displaystyle x\to 0$, $\displaystyle u\to 0$ as well.

So our limit transforms from $\displaystyle 2\lim_{x\to 0}\frac{\sin(2x)}{2x}$ to $\displaystyle 2\lim_{u\to 0}\frac{\sin u}{u}=2(1)=\color{red}\boxed{2}$

Does this make sense?

--Chris
• Oct 1st 2008, 11:25 PM
macgyver
yes thank you that helped a lot.

could u help me with this other one please?

f(x)=(e^-x)/(x)

I'm supposed to
a: use graphs and tables to find lim x approaching infinity f(x)
b: lim x approaching negative infinity f(x)
c: Identify all horizontal asymptotes