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

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?

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:

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

• Serious question: Why is this solution correct and mine isn't? – BlkPengu Mar 27 '18 at 11:48
• @BlkPengu You need to take in to account that $a$ can be positive or negative, so if $a\geq 0$, then your answer is correct, but if $a<0$, then the limit is $+\infty$. – F.A. Mar 27 '18 at 11:52
• I added a second part of the question. Could you look into it? – BlkPengu Mar 27 '18 at 12:31

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

• I published the second part of my question too late, sorry. Could you look over it again? – BlkPengu Mar 27 '18 at 11:49

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

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