# Limit of (sin 5x)/x

1. ### ktpr2

192
$$f(x) = -50x^2+5$$
$$g(x) = (sin 5x)/x$$
$$h(x) = x^2+5$$

I'm trying to find the limit of g(x) as x --> 0

I know that f(x) and h(x) are less than and greater than, respectively, than g(x) but I am unsure how to prove that w/o abusing the concept of infinity. How would I prove this so that i can show that

$$f(x)\leqq g(x) \leqq h(x)$$

and use the squeeze theorem to show that the limit as x --> 0 for g(x) = 5 because it's also 5 for f(x) and h(x)? Alternatively, I'm sure, is there a better way to go about this?

2. ### Data

998
$$\lim_{x \rightarrow 0} \frac{\sin (5x)}{x} = \left(\frac{d}{dx} \sin (5x)\right) \biggr |_{x =0} = 5\cos{0} = 5.$$

Of course, you have to know a bunch of things to be able to use that to start with. Another easy way, if you already know

$$\lim_{x\rightarrow 0} \frac{\sin x}{x} = 1$$

then just multiply top and bottom of your limit by $5$ and note that $x\rightarrow 0 \Longleftrightarrow 5x \rightarrow 0$.

There's a thread on a similar limit in the math section that might give you some ideas (there's a geometrically motivated proof there too, using the squeeze theorem, if you want to do it that way):