[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.

(L'hospital not allowed)

$\displaystyle 1)\lim\limits_{x\rightarrow0} \dfrac{\tan^22x\cdot\ln(1-3x)}{\cos3x\cdot\arcsin^3x}$

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.

(L'hospital not allowed)

$\displaystyle 2)\lim\limits_{x\rightarrow0} \left(\dfrac{x-3}{x+2}\right)^{3x-5}$

No idea how to solve this to be honest

(L'hospital not allowed)

$\displaystyle 3)\lim\limits_{x\rightarrow0} \dfrac{1-\cos3x}{x\tan2x}$

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:

$\displaystyle \lim_{x \to 0} \frac{\tan^2{2x}\ln(1-3x)}{\cos{3x}\sin^{-3}x} =$

$\displaystyle \lim_{x \to 0} -12 \left(\frac{\tan{2x}}{2x}\right)^2 \frac{\ln(1-3x)}{-3x} \frac{1}{\cos{3x}} \left(\frac{x}{\sin^{-1}{x}}\right)^3 = -12$.

For (2), you can just plug in x=0 giving $\displaystyle \left(\frac{-3}{2}\right)^{-5} = \left(-\frac{2}{3}\right)^5 = -\frac{32}{243}$.

For (3), you need to use the identity $\displaystyle \sin^2{u}=\frac{1-\cos{2u}}{2}$ to get

$\displaystyle \lim_{x\to 0} \frac{1-\cos{3x}}{x\tan{2x}} = $

$\displaystyle \lim_{x\to 0} \frac{2\sin^2{\frac{3}{2}x}}{x\tan{2x}}$

and then break it up like (1) to get the answer $\displaystyle \frac{9}{4}$.

- Hollywood

Re: Need help with eliminating indeterminate form.

Thanks.

$\displaystyle \lim_{x \to 0} -12 \left(\frac{\tan{2x}}{2x}\right)^2 \frac{\ln(1-3x)}{-3x} \frac{1}{\cos{3x}} \left(\frac{x}{\sin^{-1}{x}}\right)^3 = -12$.

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.

$\displaystyle \lim_{x \to 0} -12 \left(\frac{\tan{2x}}{2x}\right)^2 \frac{\ln(1-3x)}{-3x} \frac{1}{\cos{3x}} \left(\frac{x}{\sin^{-1}{x}}\right)^3 = -12$.

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:

$\displaystyle \cos(2x) = 1 - 2\sin^{2}(x)\,\, \therefore\,\,\cos(3x) = \cos\left(2*\frac{3x}{2}\right) = 1 - 2\sin^{2}\left(\frac{3x}{2}\right) $

$\displaystyle \text{ Thus: } 1 - \cos\left(2*\frac{3x}{2}\right) = 2\sin^{2}\left(\frac{3x}{2}\right)$

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