Limit Indeterminate Forms

  • #1
whatlifeforme
219
0

Homework Statement


find the limit.


Homework Equations


[itex]limit_{x->0+}[/itex] [itex](x+1)^{cotx}[/itex]




The Attempt at a Solution



this is of the form [itex]1^{∞}[/itex]

y = [itex](x+1)^{cotx}[/itex]
lny = cotx * ln(x+1)

not sure if this is correct so far.. and what to do next? somehow turn it into a fraction, perhaps?
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
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Homework Statement


find the limit.


Homework Equations


limit x->0+ [itex](x+1)^{cotx}[/itex]


The Attempt at a Solution



this is of the form [itex]1^{∞}[/itex]

y = [itex](x+1)^{cotx}[/itex]
lny = cotx * ln(x+1)

not sure if this is correct so far.. and what to do next? somehow turn it into a fraction, perhaps?

Yes, it's fine so far. And sure, turn it into a fraction so you can apply l'Hopital. If you have a*b you can turn it into a fraction by writing it as either a/(1/b) or b/(1/a). Which looks easier?
 
  • #3
whatlifeforme
219
0
Yes, it's fine so far. And sure, turn it into a fraction so you can apply l'Hopital. If you have a*b you can turn it into a fraction by writing it as either a/(1/b) or b/(1/a). Which looks easier?


[itex]\frac{cotx}{\frac{1}{ln(x+1)}}[/itex] = [itex]\frac{(1/0)}{(1/0)}[/itex]

[itex]\frac{-(cscx)^{2}}{\frac{1}{1/(x+1)}}[/itex] = [itex]\frac{1/0}{(1/(1/1)}[/itex]

or written in another form as:

[itex]\frac{-(cscx)^{2}}{\frac{1}{1/(x+1)}}[/itex] = [itex]\frac{\frac{1}{0}}{\frac{1}{\frac{1}{1}}}[/itex]



is that ∞/1 ??

am i correct so far. do i need to apply l'hopitals again, or is my answer correctly ∞.
 
Last edited:
  • #4
whatlifeforme
219
0
[itex]\frac{cotx}{\frac{1}{ln(x+1)}}[/itex] = [itex]\frac{(1/0)}{(1/0)}[/itex]

[itex]\frac{-(cscx)^{2}}{\frac{1}{1/(x+1)}}[/itex] = [itex]\frac{1/0}{(1/(1/1)}[/itex]

or written in another form as:

[itex]\frac{-(cscx)^{2}}{\frac{1}{1/(x+1)}}[/itex] = [itex]\frac{\frac{1}{0}}{\frac{1}{\frac{1}{1}}}[/itex]



is that ∞/1 ??

am i correct so far. do i need to apply l'hopitals again, or is my answer correctly ∞.

make that -∞.
 
  • #5
Dick
Science Advisor
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make that -∞.

You picked the hard way to do it and then you did it wrong. Try ln(x+1)/(1/cot(x))=ln(x+1)/tan(x). That's the easy way. Work it out that way, then look back and figure out what you did wrong.
 
  • #6
pierce15
315
2
Edit: You forgot to use the chain rule in your second step. I'd just do it over put tan(x) in the denominator.

Also, keep the limit in there. 1/0 isn't defined.
 
  • #7
whatlifeforme
219
0
[tex]\lim_{x->0+}\frac{ln(x+1)}{tanx} = \frac{0}{0}[/tex]

[tex]\lim_{x->0+}\frac{\frac{1}{x+1}}{(secx)^{2}} = \frac{1}{1}[/tex]
 
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  • #8
Dick
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[tex]\lim_{x->0+}\frac{ln(x+1)}{tanx} = \frac{0}{0}[/tex]

[tex]\lim_{x->0+}\frac{\frac{1}{x+1}}{(secx)^{2}} = \frac{1}{1}[/tex]

Hence? Conclusion for the original limit?
 
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  • #9
pierce15
315
2
[tex]\lim_{x->0+}\frac{ln(x+1)}{tanx} = \frac{0}{0}[/tex]

[tex]\lim_{x->0+}\frac{\frac{1}{x+1}}{(secx)^{2}} = \frac{1}{1}[/tex]
Mod note: I made the changes suggested below.
Instead of writing limit_{x->\infty}, you should write "\lim_{x\to\infty}"

Also, if you write "tex" instead of "itex", everything will look better. Only use "itex" for when you aren't starting a new line for math.

Going back to the beginning of the problem, you wrote that ln(y) was equal to the expression you just derived. So what is y?
 
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  • #10
whatlifeforme
219
0
Instead of writing limit_{x->\infty}, you should write "\lim_{x\to\infty}"

Also, if you write "tex" instead of "itex", everything will look better. Only use "itex" for when you aren't starting a new line for math.

Going back to the beginning of the problem, you wrote that ln(y) was equal to the expression you just derived. So what is y?


y = [itex](x+1)^{cotx}[/itex]
lny = cotx * ln(x+1)


[itex]\lim_{x\to0+} (x+1)^{cotx}= \lim_{x\to0+} f(x) = \lim_{x\to0+} e^{lny}[/itex]

[itex]= e^{1} [/itex]
 
  • #11
Dick
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y = [itex](x+1)^{cotx}[/itex]
lny = cotx * ln(x+1)


[itex]\lim_{x\to0+} (x+1)^{cotx}= \lim_{x\to0+} f(x) = \lim_{x\to0+} e^{lny}[/itex]

[itex]= e^{1} [/itex]

Right.
 
  • #12
pierce15
315
2
y = [itex](x+1)^{cotx}[/itex]
lny = cotx * ln(x+1)


[itex]\lim_{x\to0+} (x+1)^{cotx}= \lim_{x\to0+} f(x) = \lim_{x\to0+} e^{lny}[/itex]

[itex]= e^{1} [/itex]

Looks good. Instead of writing "itex", though, it'll look better with "tex"
 
  • #13
SithsNGiggles
186
0
Looks good. Instead of writing "itex", though, it'll look better with "tex"

You could also just use "##" at the beginning and end of whatever you want to type in latex.
 
  • #14
36,318
8,285
You could also just use "##" at the beginning and end of whatever you want to type in latex.

I would add to the OP that you don't need to have flocks of itex or tex tags - one at the beginning and the closing tag of that type at the end of the line.

## at beginning and end - same as itex at beginning and /itex at end.
$$ at beginning and end - same as tex at beginning and /tex at end.
 
  • #15
whatlifeforme
219
0
y = [tex](x+1)^{cotx}[/tex]
lny = cotx * ln(x+1)


[tex]\lim_{x\to0+} (x+1)^{cotx}= \lim_{x\to0+} f(x) = \lim_{x\to0+} e^{lny}[/tex]

[tex]= e^{1}[/tex]


**note:updated with tex.
 
  • #16
pierce15
315
2
**note:updated with tex.

Good. Also, if you want the + to be a subscript, just use ^, as follows:

[tex] \lim_{x \to 0^+ } [/tex]
 

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