[Limits] Need help with eliminating indeterminate form.

Hello again.

I need to eliminate some uncertainties. I know decent amount of trigonometric/algebraic but the reason I cant see what and how to use is probably that I haven't seen example usages of these enough. Here are some limit's I' would like to lean to calculate or to eliminate indeterminate form to be more exact.

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I tried cos(3x)= cos^3(x)- 3sin^2(x)cos(x) and tan(2x)= 2tan(x)/(1- tan^2(x)). But they didnt help. Need some other ideas.

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No idea how to solve this to be honest

(L'hospital not allowed)

I tried cos(3x)= cos^3(x)- 3sin^2(x)cos(x) and tan(2x)= 2tan(x)/(1- tan^2(x)). But they didnt help. Need some other ideas.

Re: Need help with eliminating indeterminate form.

For (1), you can break up the limit into parts:

.

For (2), you can just plug in x=0 giving .

For (3), you need to use the identity to get

and then break it up like (1) to get the answer .

- Hollywood

Re: Need help with eliminating indeterminate form.

Thanks.

.

Where did -12 come from?

That's nice that you found that trigonometric formula for 3. I sill didn't understand how exactly did you get 2 sin^2 3/2 x.

Re: Need help with eliminating indeterminate form.

Quote:

Originally Posted by

**rain1** Thanks.

.

Where did -12 come from?

That's nice that you found that trigonometric formula for 3. I sill didn't understand how exactly did you get 2 sin^2 3/2 x.

We know by trigonometry that:

Hope it helps.

Re: Need help with eliminating indeterminate form.

On #1, when I split up the expression, I did it so that each fraction has a limit of 1. So in the first fraction, I have 2x in the denominator and in the second fraction I have -3x in the denominator.

The x's cancel out: x^2 in the denominator, x in the denominator, nothing, and x^3 in the numerator.

I had to multiply by -12 to get the constants to cancel out: they were: 2^2 in the denominator, -3 in the denominator, nothing, and nothing. So -12 = 2^2 * (-3).

Hope that helps.

- Hollywood