Solving limits

Elize

I asked some question about solving limits without derivatives and I solved this question using the method I learned from those questions I asked, but the sign of answer is different from what I have found can you please check my steps and see where I did wrong?

idontknow

The mistake is at $$\displaystyle \sqrt{sin^2 u}$$

v8archie

Math Team
The square root of $\sin^2 u$ is, by definition, always positive. But for $u < 0$, as required for your limit as $u \to 0^-$, $\sin u$ is negative.

When doing substitutions you should always write down explicitly what the substitution is - it makes it easier for everyone to follow. It's also useful to try and make them so that your limits are right-handed because negative numbers frequently cause this sort of confusion.

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