- #1

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[itex]lim_{x\rightarrow 0} (\frac{\tan x}{x})^{1/x^2}[/itex]

L'Hospital's rule gets really messy, and I cannot find a suitable

choice of functions to apply the squeeze theorem. Does anyone

have any suggestions? Thank you.

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- Thread starter namu
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- #1

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[itex]lim_{x\rightarrow 0} (\frac{\tan x}{x})^{1/x^2}[/itex]

L'Hospital's rule gets really messy, and I cannot find a suitable

choice of functions to apply the squeeze theorem. Does anyone

have any suggestions? Thank you.

- #2

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Compute the natural logarithm of the limit with the help of l'Ho^pital's rule.

- #3

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Compute the natural logarithm of the limit with the help of l'Ho^pital's rule.

Thank you. That makes life so much easier. I forgot about that little trick.

I got the answer.

- #4

Mark44

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It shouldn't, because L'Hopital's Rule doesn't apply here. It applies to quotients of functions -- f(x)/g(x) -- where both functions are approaching zero or both are approaching infinity.How do I find the following limit?

[itex]lim_{x\rightarrow 0} (\frac{\tan x}{x})^{1/x^2}[/itex]

L'Hospital's rule gets really messy

And you can't simply use L'Hopital's Rule on the part being raised to the power, because the limit variable occurs in the exponent.

, and I cannot find a suitable

choice of functions to apply the squeeze theorem. Does anyone

have any suggestions? Thank you.

- #5

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It shouldn't, because L'Hopital's Rule doesn't apply here. It applies to quotients of functions -- f(x)/g(x) -- where both functions are approaching zero or both are approaching infinity.

And you can't simply use L'Hopital's Rule on the part being raised to the power, because the limit variable occurs in the exponent.

That's true, so we have

[itex]

\stackrel{lim}{x \to 0} \frac{\tan x}{x}^{1/x^2}= e^{\stackrel{lim}{x \to 0}\frac{ln \frac{\tan x}{x}}{\frac{1}{x^2} }}

[/itex]

Now,

[itex]

\stackrel{lim}{x \to 0} \frac{\tan x}{x}=1

[/itex]

hence,

[itex]

ln 1=0

[/itex]

and

[itex]

\stackrel{lim}{x \to 0} \frac{1}{x^2}=0

[/itex]

Therefore we can use l'Hospital's Rule here with [itex]f=ln \frac{\tan x}{x} [/itex] and [itex]g=1/x^2 [/itex]

- #6

Mark44

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[tex]\lim_{x \to 0}\frac{1}{x^2} \neq 0![/tex]

When you work this problem, don't include the limit until you're ready to take the limit.

Write y = (tan(x)/x)^(1/x^2)

Now take ln of both sides, and you'll have a quotient.

Take the limit of ln y and of what you have on the right, and use L'Hopital's Rule on that. This is what dextercioby was saying.

Keep in mind that the limit you'll get is the limit of the ln of something, so you'll need to adjust for that.

LaTeX tips

1) For complicated fractions, use [ tex ] tags instead of [ itex ] tags.

2) For limits, don't use \stackrel -- use \lim, like this:

[ tex ]\lim_{x \to {\infty}\frac{f(x)}{g(x)}[ /tex ]

- #7

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The correct response should be

[tex]\lim_{x \to {\infty}} \left ( \frac{\tan x}{x} \right ) ^{1/x^2}=e^{\lim_{x \to {\infty}} \frac{1}{x^2} ln \frac{\tan x}{x} } [/tex]

So,

[tex] f(x)=ln \frac{\tan x}{x} [/tex]

[tex] g(x)=x^2 [/tex]

Now [itex] f(x) \to 0 [/itex] and [itex] g(x) \to 0 [/itex]

- #8

Dick

Science Advisor

Homework Helper

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The correct response should be

[tex]\lim_{x \to {\infty}} \left ( \frac{\tan x}{x} \right ) ^{1/x^2}=e^{\lim_{x \to {\infty}} \frac{1}{x^2} ln \frac{\tan x}{x} } [/tex]

So,

[tex] f(x)=ln \frac{\tan x}{x} [/tex]

[tex] g(x)=x^2 [/tex]

Now [itex] f(x) \to 0 [/itex] and [itex] g(x) \to 0 [/itex]

Ok, so now you have a 0/0 limit. What do you get for the limit using l'Hopital? This whole thing is probably worked out easier using taylor series, if you know that approach.

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