Can someone double-check my work here? Thanks, if you can. (Nod)

Put x to the denominator to use L'Hopital's rule, etc. Applied chain rules, etc.

Thus, if x goes to infinity, the above is sin(0) = 0. Final Answer: Zero

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- Jan 21st 2010, 06:28 PMLord DarkinDouble-check this limit please
Can someone double-check my work here? Thanks, if you can. (Nod)

Put x to the denominator to use L'Hopital's rule, etc. Applied chain rules, etc.

Thus, if x goes to infinity, the above is sin(0) = 0. Final Answer: Zero - Jan 21st 2010, 06:35 PMKrizalid
Yes, it's okay, but you don't need that rule, you can turn that limit into a known one with a simple substitution.

- Jan 22nd 2010, 12:39 PMLord Darkin
Oh, really? What is this substitution? :)

- Jan 22nd 2010, 12:39 PMKrizalid
- Jan 23rd 2010, 03:58 AMSeulementrien
tx=1

so itequals to lim [(1-cost)/t]

t->0