Science:Math Exam Resources/Courses/MATH110/April 2011/Question 02 (b)

MATH110 April 2011

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Question 02 (b)
Evaluate the following limit, if it exists. $\displaystyle \lim _{x\to 0^{+}}{\frac {\ln(\sin x)}{\ln(\cos x)}}$

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Try evaluating the limit directly. Pay attention to the directional limit - the fact that this is a right hand limit will impact the sign of your answer.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. We can evaluate this limit directly. We start by evaluating what happens to $\sin x$ and $\cos x$ as $x\to 0^{+}$. Because this is a directional limit from the right, we will have $\sin x\to 0^{+}$ (on a sine graph, as you approach x = 0 from the right, the y value is approaching zero from above, hence approaching $0^{+}$) and $\cos x\to 1^{-}$ (similarly, approaching x = 0 from the right on a cosine graph means approaching y = 1 from below, hence $1^{-}$). Replacing $\sin x$ with y and $\cos x$ with z, we can thus rewrite the limit as a fraction of the limits: $\lim _{x\to 0^{+}}{\frac {\ln(\sin x)}{\ln(\cos x)}}={\frac {\lim _{y\to 0^{+}}\ln(y)}{\lim _{z\to 1^{-}}\ln(z)}}$ Again, the fact that this is a directional limit means that for the first limit, as $y\to 0^{+}$, $\ln(y)\to -\infty$ and for the second limit, as $z\to 1^{-}$, $\ln(z)\to 0^{-}$. So we have: ${\frac {\lim _{y\to 0^{+}}\ln(y)}{\lim _{z\to 1^{-}}\ln(z)}}={\frac {-\infty }{0^{-}}}$ Recall that when a function goes to zero from the left in the denominator, this means that we eventually divide by a very small negative value (for example, -1/100000). Using fraction rules, this very small value in the denominator is the same as multiplying by a very large negative value (in my example, -100000) in the numerator. Thus going to zero in the denominator corresponds to multiplying the numerator by a very large number. Since our numerator is already going to negative infinity, the fact that we have $0^{-}$ in the denominator will make the value of the limit greater, and turn it back from negative to positive. Thus the value of this limit is $+\infty$, $\displaystyle \lim _{x\to 0^{+}}{\frac {\ln(\sin x)}{\ln(\cos x)}}=+\infty$

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