Compute the following limit (Use L'Hôpital's rule): [lncos(a/x)]/x^(-2) as x tends to infinity.
Last edited by mr fantastic; May 6th 2010 at 06:05 PM. Reason: Copied from title into main body of post.
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Originally Posted by feyomi [lncos(a/x)]/x^(-2) as x tends to infinity. Hi feyomi, You can now evaluate the limit, since as x goes to infinity hence the limit is
Originally Posted by feyomi [lncos(a/x)]/x^(-2) as x tends to infinity. This is a bit hard to read... Is it ?
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