Science:Math Exam Resources/Courses/MATH100/December 2019/Question 1 (b)

MATH100 December 2019

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Question 1 (b)
Evaluate $\lim _{x\to 0}{\frac {\log(\cos(x))}{\log \left(\cos({\frac {x}{2}})\right)}}$. (Here log stands for the natural logarithm.)

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 hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
Use the trigonometric identity $\cos(x/2)={\sqrt {\frac {1+\cos(x)}{2}}}$.

Hint 2
Use L'Hôpital after the using the trig identity.

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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. The limit can be solved by using the trigonometric identity $\cos(x/2)={\sqrt {\frac {1+\cos(x)}{2}}}$ and L'Hôpital's rule. Using the trig identity we get: $\lim _{x\to 0}{\frac {\log(\cos(x))}{\log \left(\cos({\frac {x}{2}})\right)}}=\lim _{x\to 0}{\frac {\log(\cos(x))}{{\frac {1}{2}}(\log(1+\cos(x))+\log(2))}}.$ Here the derivative of the numerator above is ${\frac {d}{dx}}(\log(\cos(x)))=-{\frac {\sin(x)}{\cos(x)}}$ and the derivative of the denominator yields $-{\frac {\sin(x)}{2(1+\cos(x))}}$. It follows that the functions satisfy all requirements for L'Hôpital's rule to apply. Therefore: $\lim _{x\to 0}{\frac {\log(\cos(x))}{\log \left(\cos({\frac {x}{2}})\right)}}=\lim _{x\to 0}{\frac {-{\frac {\sin(x)}{\cos(x)}}}{-{\frac {\sin(x)}{2(1+\cos(x))}}}}=\lim _{x\to 0}{\frac {2(1+\cos(x))}{\cos(x)}}={\frac {2(1+\cos(0))}{\cos(0)}}=4.$