Question

How to calculate this limit?
http://imgur.com/a/x8a93

Answer 1

hey are you still there?

Answer 2

so you have to do a little algebra first

Answer 3

I would use a Taylor Expansion here. Sure, you can twist it into a L'Hopital rule, but it twists itself back out again and with trig functions you can find yourself going round in circles.

So if we agree that the cosine Taylor Expansion is:

\(\cos x = 1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} + \mathcal O (x^6)\)

and if we add the components of the limit such that:

\(\lim\limits_{x \to 0} \left( \dfrac{1}{\cos x - 1} - \dfrac {1}{x} \right)\)

\(= \lim\limits_{x \to 0} \dfrac{x - \cos x + 1}{x( \cos x - 1)}\)

We can insert the series, going overboard with the higher terms, just to be sure....:

\(= \lim\limits_{x \to 0} \dfrac{x - (1 - \dfrac{x^2}{2!} + \mathcal O (x^4)) + 1}{x \left(1 - \dfrac{x^2}{2!} + \mathcal O (x^4) - 1 \right) } \)

and then some of that "little algebra" :-))

\(= \lim\limits_{x \to 0} \dfrac{x + \dfrac{x^2}{2!} + \mathcal O (x^4)) }{x \left( - \dfrac{x^2}{2!} + \mathcal O (x^4) \right) }\)

\(= - \lim\limits_{x \to 0} \dfrac{1 + \dfrac{x}{2!} + \mathcal O (x^3)) }{ \left( \dfrac{x^2}{2!} + \mathcal O (x^4) \right) }\)

\(= - \lim\limits_{x \to 0} \dfrac{1 + \mathcal O (x) }{ \mathcal O (x^2) } = - \infty \)

So it's a proper 2-sided limit as well, as the denominator is sign neutral

:-)

Now having done it that way, i think that putting the expansion in right at the start might have been the real smart thing to do.

Answer 4

Thanks a lot, Irishboy123.

Answer 5

mp

:)