Solving limits

Feb 2018
63
3
Iran

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?
 
Dec 2015
1,084
169
Earth
The mistake is at \(\displaystyle \sqrt{sin^2 u}\)
 

v8archie

Math Team
Dec 2013
7,713
2,682
Colombia
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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