Hi,

I have

Now replace the trig functions with its corresponding taylor polynomial

This limit evaluates to infinity, i don't understand why. It looks more like -2 to me

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- Jan 2nd 2010, 03:26 PMJonesI don't understand this limit problem
Hi,

I have

Now replace the trig functions with its corresponding taylor polynomial

This limit evaluates to infinity, i don't understand why. It looks more like -2 to me - Jan 2nd 2010, 04:15 PMANDS!
Too much work (unless you are required to use Taylor representations), both functions approach zero, so you can easily just use L'hopitals rule.

- Jan 2nd 2010, 08:02 PMtonio
- Jan 2nd 2010, 11:11 PMmatheagle
or you can use L'Hopital's Rule here, which I just noticed was also recommended.

- Jan 2nd 2010, 11:21 PMDrexel28
Another convoluted way to show the limit doesn't exist is to assume it does. Then . But . And since , we may by assumption conclude that which tells us that exists, but this is ludicrous since and . Thus, our assumption was false and the original limit does not exist.