# Limit Indeterminate Forms

find the limit.

## Homework Equations

$limit_{x->0+}$ $(x+1)^{cotx}$

## The Attempt at a Solution

this is of the form $1^{∞}$

y = $(x+1)^{cotx}$
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?

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Dick
Homework Helper

find the limit.

## Homework Equations

limit x->0+ $(x+1)^{cotx}$

## The Attempt at a Solution

this is of the form $1^{∞}$

y = $(x+1)^{cotx}$
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?

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?

$\frac{cotx}{\frac{1}{ln(x+1)}}$ = $\frac{(1/0)}{(1/0)}$

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

or written in another form as:

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

is that ∞/1 ??

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

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$\frac{cotx}{\frac{1}{ln(x+1)}}$ = $\frac{(1/0)}{(1/0)}$

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

or written in another form as:

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

is that ∞/1 ??

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

Dick
Homework Helper
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.

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.

$$\lim_{x->0+}\frac{ln(x+1)}{tanx} = \frac{0}{0}$$

$$\lim_{x->0+}\frac{\frac{1}{x+1}}{(secx)^{2}} = \frac{1}{1}$$

Last edited by a moderator:
Dick
Homework Helper
$$\lim_{x->0+}\frac{ln(x+1)}{tanx} = \frac{0}{0}$$

$$\lim_{x->0+}\frac{\frac{1}{x+1}}{(secx)^{2}} = \frac{1}{1}$$
Hence? Conclusion for the original limit?

Last edited by a moderator:
$$\lim_{x->0+}\frac{ln(x+1)}{tanx} = \frac{0}{0}$$

$$\lim_{x->0+}\frac{\frac{1}{x+1}}{(secx)^{2}} = \frac{1}{1}$$
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:
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 = $(x+1)^{cotx}$
lny = cotx * ln(x+1)

$\lim_{x\to0+} (x+1)^{cotx}= \lim_{x\to0+} f(x) = \lim_{x\to0+} e^{lny}$

$= e^{1}$

Dick
Homework Helper
y = $(x+1)^{cotx}$
lny = cotx * ln(x+1)

$\lim_{x\to0+} (x+1)^{cotx}= \lim_{x\to0+} f(x) = \lim_{x\to0+} e^{lny}$

$= e^{1}$
Right.

y = $(x+1)^{cotx}$
lny = cotx * ln(x+1)

$\lim_{x\to0+} (x+1)^{cotx}= \lim_{x\to0+} f(x) = \lim_{x\to0+} e^{lny}$

$= e^{1}$
Looks good. Instead of writing "itex", though, it'll look better with "tex"

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.

Mark44
Mentor
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.

y = $$(x+1)^{cotx}$$
lny = cotx * ln(x+1)

$$\lim_{x\to0+} (x+1)^{cotx}= \lim_{x\to0+} f(x) = \lim_{x\to0+} e^{lny}$$

$$= e^{1}$$

**note:updated with tex.

**note:updated with tex.
Good. Also, if you want the + to be a subscript, just use ^, as follows:

$$\lim_{x \to 0^+ }$$