How Do You Solve the Limit of (x+1)^(cotx) as x Approaches 0+?

[whatlifeforme]

Homework Statement

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?

 

[Dick]

Homework Statement

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?

 

[whatlifeforme]

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 ∞.

 

[whatlifeforme]

\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]

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.

 

[pierce15]

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.

 

[whatlifeforme]

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

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

 

[Dick]

\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?

 

[pierce15]

\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.[/color]
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?

 

[whatlifeforme]

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]

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.

 

[pierce15]

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"

 

[SithsNGiggles]

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]

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.

 

[whatlifeforme]

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.

 

[pierce15]

**note:updated with tex.

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

\lim_{x \to 0^+ }