One-Sided Limits - Step by Step Guide

[ae4jm]

[SOLVED] One-Sided Limits

Homework Statement

I've been having problems solving for these one-sided limits.

1.$$\stackrel{lim}{x\rightarrow\pi^-}\frac{sinx}{1-cosx}$$

2.$$\stackrel{lim}{x\rightarrow\O^+}xln(x)$$

3.$$\stackrel{lim}{x\rightarrow\frac{\pi}{2}^-}(sec (x)-tan(x))$$

Homework Equations

The Attempt at a Solution

I solved #1 as =0/1 =0



I solved #2 as $$\stackrel{lim}{x\rightarrow\O^+}\frac{ln(x)}{\frac{1}{x}}$$

=$$\frac{\frac{1}{x}}{1}$$ = $$\frac{\frac{1}{0}}{1}$$

=0


I solved #3 as $$\stackrel{lim}{x\rightarrow\frac{\pi}{2}^-}(sec (x)-tan(x))$$

=$$\stackrel{lim}{x\rightarrow\frac{\pi}{2}^-}(\frac{1}{cosx}-\frac{sinx}{cosx})$$


=$$\stackrel{lim}{x\rightarrow\frac{\pi}{2}^-}(\frac{1-sinx}{cosx})$$ and I solved this down to =$$\stackrel{lim}{x\rightarrow\frac{\pi}{2}^-}tan(x)$$


= undefined

I'm missing some steps in these problems. Any help from you is greatly appreciated!

 

[HallsofIvy]

Homework Statement

I've been having problems solving for these one-sided limits.

1.$$\stackrel{lim}{x\rightarrow\pi^-}\frac{sinx}{1-cosx}$$

2.$$\stackrel{lim}{x\rightarrow\O^+}xln(x)$$

3.$$\stackrel{lim}{x\rightarrow\frac{\pi}{2}^-}(sec (x)-tan(x))$$

Homework Equations

The Attempt at a Solution

I solved #1 as =0/1 =0

How did you get 0/1? What is cos($\pi$)?

I solved #2 as $$\stackrel{lim}{x\rightarrow\O^+}\frac{ln(x)}{\frac{1}{x}}$$

=$$\frac{\frac{1}{x}}{1}$$ = $$\frac{\frac{1}{0}}{1}$$

= 0

and why would 1/0 = 0??

I solved #3 as $$\stackrel{lim}{x\rightarrow\frac{\pi}{2}^-}(sec (x)-tan(x))$$

=$$\stackrel{lim}{x\rightarrow\frac{\pi}{2}^-}(\frac{1}{cosx}-\frac{sinx}{cosx})$$


=$$\stackrel{lim}{x\rightarrow\frac{\pi}{2}^-}(\frac{1-sinx}{cosx})$$ and I solved this down to =$$\stackrel{lim}{x\rightarrow\frac{\pi}{2}^-}tan(x)$$


= undefined

Since you used L'Hopital's rule in #2, why not here?
[tex]\lim_{x\rightarrow \frac{\pi^-}{2}} \frac{1- sin(x)}{cos(x)}= \lim_{x\rightarrow \frac{\pi^-}{2}}\frac{-cos(x)}{sin(x)}[/itex]

I'm missing some steps in these problems. Any help from you is greatly appreciated!

By the way- don't use "stackrel" in that way.
$$\lim_{x\rightarrow \frac{\pi^-}{2}}$$
is much easier to read than
$$\stackrel{lim}{x\rightarrow\frac{\pi}{2}^-}$$

 

[HallsofIvy]

Homework Statement

I've been having problems solving for these one-sided limits.

1.$$\stackrel{lim}{x\rightarrow\pi^-}\frac{sinx}{1-cosx}$$

2.$$\stackrel{lim}{x\rightarrow\O^+}xln(x)$$

3.$$\stackrel{lim}{x\rightarrow\frac{\pi}{2}^-}(sec (x)-tan(x))$$

Homework Equations

The Attempt at a Solution

I solved #1 as =0/1 =0



I solved #2 as $$\stackrel{lim}{x\rightarrow\O^+}\frac{ln(x)}{\frac{1}{x}}$$

=$$\frac{\frac{1}{x}}{1}$$ = $$\frac{\frac{1}{0}}{1}$$

=0

And why would 1/0 be 0?

I solved #3 as $$\stackrel{lim}{x\rightarrow\frac{\pi}{2}^-}(sec (x)-tan(x))$$

=$$\stackrel{lim}{x\rightarrow\frac{\pi}{2}^-}(\frac{1}{cosx}-\frac{sinx}{cosx})$$


=$$\stackrel{lim}{x\rightarrow\frac{\pi}{2}^-}(\frac{1-sinx}{cosx})$$ and I solved this down to =$$\stackrel{lim}{x\rightarrow\frac{\pi}{2}^-}tan(x)$$

= undefined

Since you used L'Hopital's rule in #2, why not here?
$$\lim_{x\rightarrow\frac{\pi^-}{2}}\frac{1- sin x}{cos x}= \lim_{x\rightarrow\frac{\pi^-}2}}\frac{cos x}{sin x}$$
What is that limit?

I'm missing some steps in these problems. Any help from you is greatly appreciated!

By the way "stackrel" makes that very hard to read.
$$\lim_{x\rightarrow\frac{\pi^-}{2}}$$
is much better than
$$\stackrel{lim}{x\rightarrow\frac{\pi}{2}^-}$$
Click on the LaTex to see the code.

 

[ae4jm]

#1. I should have gotten 0/2 for #1, because cos(pi)=1 and 0/(1-(-1))=0/2
I thought that 0/(any nymber greater than or less than 0) is equal to 0, but if the fraction was 2/0, there would be an infinite, positive for the limit approaching from the right and negative infinity for numbers approaching from the left?

#2. I reworked through the problem several times and I keep getting (1/0)/1 which is, after going over my notes several times, = to infinity/1 which is = infinity

#3. I got = 0, after following L'Hopital's as you said. I originally -0/1, which would be = 0.

I apologize about the laytex from my previous post. I was unable to check the box for quoting your reply and when I clicked to preview my post on this reply I wasn't seeing any latex images, so I just used the harder to read way of expressing my answers--sorry for not using latex.

EDIT:
Is this what you were trying to get me to understand? That 1/0 in limits is equal to an infinite answer and that 0/1 in limits is equal to 0?

 

[ae4jm]

Am I going about the solutions correctly now?

 

[ae4jm]

I found my error on problem #2. It should be equal to 0 after rewriting the form of the limit because it was indeterminant. I got it finalized and came up with 0 for the answer to number 2. Thanks HallsofIvy.