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One-Sided Limits

  • Thread starter ae4jm
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[SOLVED] One-Sided Limits

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

1.[tex]\stackrel{lim}{x\rightarrow\pi^-}\frac{sinx}{1-cosx}[/tex]

2.[tex]\stackrel{lim}{x\rightarrow\O^+}xln(x)[/tex]

3.[tex]\stackrel{lim}{x\rightarrow\frac{\pi}{2}^-}(sec (x)-tan(x))[/tex]



2. Homework Equations



3. The Attempt at a Solution

I solved #1 as =0/1 =0



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

=[tex]\frac{\frac{1}{x}}{1}[/tex] = [tex]\frac{\frac{1}{0}}{1}[/tex]

=0


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

=[tex]\stackrel{lim}{x\rightarrow\frac{\pi}{2}^-}(\frac{1}{cosx}-\frac{sinx}{cosx})[/tex]


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


= undefined

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

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
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1. Homework Statement
I've been having problems solving for these one-sided limits.

1.[tex]\stackrel{lim}{x\rightarrow\pi^-}\frac{sinx}{1-cosx}[/tex]

2.[tex]\stackrel{lim}{x\rightarrow\O^+}xln(x)[/tex]

3.[tex]\stackrel{lim}{x\rightarrow\frac{\pi}{2}^-}(sec (x)-tan(x))[/tex]



2. Homework Equations



3. The Attempt at a Solution

I solved #1 as =0/1 =0
How did you get 0/1? What is cos([itex]\pi[/itex])?


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

=[tex]\frac{\frac{1}{x}}{1}[/tex] = [tex]\frac{\frac{1}{0}}{1}[/tex]

= 0
and why would 1/0 = 0??


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

=[tex]\stackrel{lim}{x\rightarrow\frac{\pi}{2}^-}(\frac{1}{cosx}-\frac{sinx}{cosx})[/tex]


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


= 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.
[tex]\lim_{x\rightarrow \frac{\pi^-}{2}}[/tex]
is much easier to read than
[tex]\stackrel{lim}{x\rightarrow\frac{\pi}{2}^-}[/tex]
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
41,738
899
1. Homework Statement
I've been having problems solving for these one-sided limits.

1.[tex]\stackrel{lim}{x\rightarrow\pi^-}\frac{sinx}{1-cosx}[/tex]

2.[tex]\stackrel{lim}{x\rightarrow\O^+}xln(x)[/tex]

3.[tex]\stackrel{lim}{x\rightarrow\frac{\pi}{2}^-}(sec (x)-tan(x))[/tex]



2. Homework Equations



3. The Attempt at a Solution

I solved #1 as =0/1 =0



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

=[tex]\frac{\frac{1}{x}}{1}[/tex] = [tex]\frac{\frac{1}{0}}{1}[/tex]

=0
And why would 1/0 be 0?


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

=[tex]\stackrel{lim}{x\rightarrow\frac{\pi}{2}^-}(\frac{1}{cosx}-\frac{sinx}{cosx})[/tex]


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

= 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}[/tex]
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.
[tex]\lim_{x\rightarrow\frac{\pi^-}{2}}[/tex]
is much better than
[tex]\stackrel{lim}{x\rightarrow\frac{\pi}{2}^-}[/tex]
Click on the LaTex to see the code.
 
  • #4
79
0
#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?
 
Last edited:
  • #5
79
0
Am I going about the solutions correctly now?
 
  • #6
79
0
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.
 

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