• Support PF! Buy your school textbooks, materials and every day products Here!

Limit Indeterminate Forms

  • #1

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
26,258
618

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
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
[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
Homework Helper
26,258
618
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
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
[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]
 
Last edited by a moderator:
  • #8
Dick
Science Advisor
Homework Helper
26,258
618
[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?
 
Last edited by a moderator:
  • #9
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?
 
Last edited by a moderator:
  • #10
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
Science Advisor
Homework Helper
26,258
618
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
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
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
33,638
5,300
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
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
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]
 

Related Threads on Limit Indeterminate Forms

  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
1
Views
557
Replies
1
Views
1K
Replies
22
Views
4K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
24
Views
215
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
3
Views
600
  • Last Post
Replies
3
Views
2K
Top