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How do I do this problem? Find the limit: lim as x --> infinity of x * sin (1/x)=
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let u=1/x as x->infinity, u->0
limit turn out to be "1"
i doubt it
oh maybe i am wrong
\[ L=\lim_{u\to0}{\sin u\over u}=1 \]
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yeah, i am wrong
@satellite73 :)
to err is human
How'd you get to the L equation?
\[u={1\over x}\\as\;x\to\infty,\;u\to0 L=\lim_{x\to\infty}x\sin\left({1\over x}\right)\\ L=\lim_{u\to0}{\sin u\over u}=1\]
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Man I feel like an idiot, I still don't get it... Why did you put the sin u over u?
by the substitution
1/x =u and x=1/u
you see it now?
I guess so man... I appreciate your help, I hope this isnt on the exam lol.
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