I'm dumbfounded here... Limits 0/0

[stalker7d7]

Two limit problems I can not figure out for the life of me... I've tried L'Hospitals rules, I've tried simplifying. I just can't get it NOT to be a 0/0, or to match up correctly to the correct limit.

Lim as x->0+ x^2 / (cos(x)-1)



Figrued this out finally *Lim as x_>2 (4x-4-x^2) / (x^3-2x^2-4x+8)

Tried all the wrong solutions. Got it by factoring the bottom poly, using (4x-4-x^2) * (-x-2)


Any help is appreciated!

 

[soroban]

Hello, stalker7d7!

I don't understand your difficulty with the first one . . .

\(\displaystyle \displaystyle\lim_{x\to0^+}\frac{x^2}{\cos x - 1}\)

\(\displaystyle \text{Since }\dfrac{0^2}{\cos 0 - 1} \:=\:\dfrac{0}{0}\text{, we can apply L'Hopital.}\)


\(\displaystyle \displaystyle \text{And we have: }\:\lim_{x\to0}\frac{2x}{\text{-}\sin x} \;\;\Rightarrow\;\;\frac{0}{0}\)


\(\displaystyle \displaystyle\text{Apply L'Hopital again: }\:\lim_{x\to0}\frac{2}{\text{-}\cos x} \;=\;\frac{2}{\text{-}1} \;=\;-2\)

\(\displaystyle \displaystyle \lim_{x\to2} \frac{4x-4-x^2}{x^3-2x^2-4x+8}\)

We have: .\(\displaystyle \dfrac{-(x^2-4x+4)}{x^2(x-2) - 4(x-2)} \;=\; \dfrac{-(x-2)^2}{(x-2)(x^2-4)} \)

. . . . . . . . \(\displaystyle =\;\dfrac{-(x-2)(x-2)}{(x-2)(x-2)(x+2)} \;=\;\dfrac{-1}{x+2}\)


\(\displaystyle \displaystyle\text{Therefore: }\:\lim_{x\to2}\frac{-1}{x+2} \;=\;-\frac{1}{4}\)

 

[stalker7d7]

Oh. I didn't realize you could apply L'Hospitals rule indefinately (as long as it meets requirements), or rather, I didn't even think of it. Makes perfect sense now.

Thanks a lot!