Trouble with l'Hôpital's rule for $\lim_{x\rightarrow 0} \frac{4x+4\sin x}{10x+10\cos x}$

Solution 1:

Original posted question:

$$\lim_{x\rightarrow 0} \frac{4x+4\sin x}{10x+10\sin x} = \lim_{x\to 0} \frac{(4x + 4\sin x)'}{(10x + 10 \sin x)'} = \lim_{x\to 0} \frac{4+4\cos x}{10 + 10\cos x} = \frac {4 + 4}{10 + 10} = \frac {8}{20} = \frac 25$$

Since you meant to post $$\lim_{x\rightarrow 0} \frac{4x+4\sin x}{10x+10\cos x}$$ note that in this case, l'Hôpital is not applicable (the limit does not at first evaluate to an indeterminate limit). Nor would we want to use it! It is easily solved by evaluating immediately:

$$\lim_{x\rightarrow 0} \frac{4x+4\sin x}{10x+10\cos x} = \frac{ 0 + 0}{0 + 10(1)} = \frac {0}{10} = 0$$

IMPORTANT TO REMEMBER: We apply l'Hôpital's rule if and only if a limit evaluates to an indeterminate form. That bold-face link will take you to Wikipedia's concise list of "what counts" as an indeterminate form, and why.

Solution 2:

Just plug in! The denominator is $10$, not zero. ;)