Calculate Limit by applying L'Hôpital's Rule and Taylor to $\lim\limits_{x \to 0} f(x) = \lim\limits_{x \to 0} \frac{ax -\cos(ax)} {ax^2} $

Question

My problem is that I'm thinking I'm supposed to use L'Hôpital on $f(x)$ but I don't get how this is supposed to be possible with the numerator converging to $-1$.

What I'd idealy want is something like $\sin(ax)$ to converge to $0$.

$$\lim\limits_{x \to 0} f(x) = \lim\limits_{x \to 0} \frac{ax -\cos(ax)} {ax^2} = \frac {-1} 0 = -\infty $$

Am I doing it wrong/ overlooking something?

I'm also supposed to use Taylor Series Approximation for

$$\cos(x) = \sum_{k=0}^\infty (-1)^k \frac{x^{2k}}{2k!}$$

I have no idea how to approach that. What would be my stating point to use Taylor?

Answer 1

It's not an indeterminate form, so you can't apply L'Hôpital. In fact, the limit is $$\lim_{x\to 0}\frac{ax - cos(ax)}{ax^2} = \lim_{x\to 0}\frac{x-\frac{cos(ax)}{a}}{x^2} = \frac{-\infty}{a}$$

If you want to use Taylor series, as $x \to 0$ then

$$\cos(x) = \sum_{k=0}^\infty (-1)^k \frac{x^{2k}}{2k!}$$

So we can use it in our limit:

$$\lim_{x\to 0}\frac{ax - cos(ax)}{ax^2} = \lim_{x\to 0}\frac{ax - \frac{1}{2} + \frac{(ax)^2}{2} - \frac{(ax)^4}{4} + o\left((ax)^6\right)}{ax^2} = \ldots$$

Answer 2

You are right ! You can not apply L'Hôpital to determine $ \lim\limits_{x \to 0} \frac{ax -\cos(ax)} {ax^2}$.

Answer 3

You can't apply l'Hospital since it is not an ideterminate form indeed, with symbolic notation

$$\frac{ax -\cos(ax)} {ax^2}\to\frac{0-1}{0^+}=-\infty$$