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Limit of (sin 5x)/x

by ktpr2
Tags: 5x or x, limit
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ktpr2
#1
Apr7-05, 07:30 PM
P: 192
[tex]f(x) = -50x^2+5[/tex]
[tex]g(x) = (sin 5x)/x[/tex]
[tex]h(x) = x^2+5[/tex]

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

[tex]f(x)\leqq g(x) \leqq h(x)[/tex]

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?
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#2
Apr7-05, 07:39 PM
P: 998
[tex] \lim_{x \rightarrow 0} \frac{\sin (5x)}{x} = \left(\frac{d}{dx} \sin (5x)\right) \biggr |_{x =0} = 5\cos{0} = 5.[/tex]

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

[tex]\lim_{x\rightarrow 0} \frac{\sin x}{x} = 1[/tex]

then just multiply top and bottom of your limit by [itex]5[/itex] and note that [itex]x\rightarrow 0 \Longleftrightarrow 5x \rightarrow 0[/itex].

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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