# Apply L'Hopital's Rule to the Following Limit

I am having trouble utilizing L'Hopital's rule on the following limit:

$$\lim\limits_{w \to 2} \frac{\sin(wt) - (w/2)(\sin(2t))}{(4-w^2)}$$ where $$t$$ is constant.

I apply L'Hopital's rule to the expression once, converting it to

$$\lim\limits_{w \to 2}\frac{(t(\cos(wt)) - (\sin(2t)/2))}{(4-2w)}$$

However, from this point forward I am confused as to how to continue utilizing L'Hopital's Rule. It is not clear to me that the numerator simplifies to $$0$$ or infinity, so I am not sure I can even keep differentiating.

I am supposed to conclude that the given limit is equal to $$(-t/4)(\cos(2t)) + (1/8)(\sin(2t))$$.

• (4-w^2)'=-2w not 4-2w Commented Aug 25, 2020 at 5:38
• its bit hard to read do you mind using MathJax math.meta.stackexchange.com/questions/5020/… Commented Aug 25, 2020 at 5:43

For the version

$$\lim_{w\to 2} \frac{(\sin(wt) - (w/2))\sin(2t)}{(4-w^2)}$$

it is not in general an indeterminate form unless

$$(\sin(wt) - (w/2))\sin(2t) \to (\sin(2t) - 1)\sin(2t)=0$$

that is

$$2t=k\pi \quad \lor\quad 2t=(2k+1)\frac \pi 2$$

then for $$2t=k\pi$$

$$\frac{(\sin(wt) - (w/2))\cdot 0}{4-w^2}=0$$

for $$2t=(2k+1)\frac \pi 2$$

$$\lim_{w\to 2}\frac{\sin(wt) - (w/2)}{4-w^2}=\lim_{w\to 2}\frac{t\cos(wt) - \frac12}{-2w} \to \frac{t\cos(2t) - \frac12}{-4}=\frac18$$

Otherwise the limit doesn't exist, right and left limits diverge depending on the sign of $$(\sin(2t) - 1)\sin(2t)$$.

For the second version we have that

$$\lim_{w\to 2} \frac{\sin(wt) - \frac w 2sin(2t)}{4-w^2}=\lim_{w\to 2} \frac{t\cos(wt) - \frac12sin(2t)}{-2w}=\frac{t\cos(2t) - \frac12\sin(2t)}{-4}$$