Science:Math Exam Resources/Courses/MATH104/December 2016/Question 07 (b)

MATH104 December 2016

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Question 07 (b)
Interpret the following limit as the slope of a tangent line to some curve $y=f(x)$ at a point $x=a$. Determine what $f(x)$ and a are and then use differentiation rules to evaluate the limit: $\lim _{x\to {\frac {\pi }{4}}}{\frac {\sin x-\cos x}{x-{\frac {\pi }{4}}}}$

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
Let $f(x)=\sin x-\cos x$. Using $f\left({\frac {\pi }{4}}\right)=0$, write the limit as a derivative.

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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. Let $f(x)=\sin x-\cos x$. Then, we can easily see that $f\left({\frac {\pi }{4}}\right)=\sin \left({\frac {\pi }{4}}\right)-\cos \left({\frac {\pi }{4}}\right)={\frac {\sqrt {2}}{2}}-{\frac {\sqrt {2}}{2}}=0$. Using this, we can write the limit as the derivative of $f$ at ${\frac {\pi }{4}}$; $\lim _{x\to {\frac {\pi }{4}}}{\frac {\sin x-\cos x}{x-{\frac {\pi }{4}}}}=\lim _{x\to {\frac {\pi }{4}}}{\frac {f(x)-f\left({\frac {\pi }{4}}\right)}{x-{\frac {\pi }{4}}}}=f'\left({\frac {\pi }{4}}\right).$ To evaluate the limit, we first find the derivative of $f$; $f'(x)=(\sin x-\cos x)'=\cos x+\sin x$ and then plug $x={\frac {\pi }{4}}$; $f'\left({\frac {\pi }{4}}\right)=\cos \left({\frac {\pi }{4}}\right)+\sin \left({\frac {\pi }{4}}\right)={\frac {\sqrt {2}}{2}}+{\frac {\sqrt {2}}{2}}={\sqrt {2}}$. To sum, we get $\lim _{x\to {\frac {\pi }{4}}}{\frac {\sin x-\cos x}{x-{\frac {\pi }{4}}}}=\color {blue}{\sqrt {2}}$.