limit of (sin 5x)/x

by ktpr2
Tags: 5x or x, limit
 P: 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?
 P: 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): http://physicsforums.com/showthread.php?t=70358

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