As the title says, I am looking for ways to estimate numerically $\lim \limits _{x \to 0} \frac {\sin 4x} x$. So far, I've tried filling in numbers on either end of zero to make an estimate, and keep getting answers around $.0697$ or $.0698$, but the homework website I am on is marking me wrong.
Am I solving this the wrong way? Does anyone know how else I should go about this? Please let me know.
On
You could also use Taylor series expansion for $\sin(4x)$ and divide by $x$ to get:
$$\frac{\sin(4x)}{x}=\frac{4x-32x^3/3+128x^5/15+\cdots}{x}$$
Clearly the limit here is $4$ as $x$ approaches zero.
You could argue that since Taylor series is not exact, then this approach using estimation or not, well, I just wanted to show you another way.
You're using degrees where you should be using radians.
$$ \lim_{x\to0}\frac{\sin x} x = \text{a nonzero number, equal to $1$ if radians are used.} $$ $$ \frac d {dx} \sin x = C\cdot\cos x,\text{ and $C=1$ if radians are used.} $$ There is a reason why radians are used in calculus, and this is it.