Science:Math Exam Resources/Courses/MATH104/December 2010/Question 01 (e)

MATH104 December 2010

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Question 01 (e)
If a function y=f(x) is differentiable at x=3 and $\displaystyle {}f'(3)=5$, find the limit $\lim _{x\to {3}}{\frac {x^{2}-3x}{f(3)-f(x)}}$

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 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
How could you possibly use the fact that $\displaystyle f'(3)=5$ to solve this limit?

Hint 2
Is there any way to use the definition of the 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. First of all, both the numerator and denominator converge to zero as $x\to 3$, so we cannot solve the limit directly. Instead, we need to make use of the fact that $\displaystyle f'(3)=5$. Taking this as a hint, let's recall the definition of the derivative as the limit of a difference quotient: $f'(a)=\lim _{x\to a}{\frac {f(x)-f(a)}{x-a}}$ For us, a = 3 and $\displaystyle f'(3)=5$, so we know the left hand side of the equation above. To find the limit of the quotient we are given, let us rewrite our quotient in a similar form as the difference quotient: ${\begin{aligned}{\frac {x^{2}-3x}{f(3)-f(x)}}&=x{\frac {x-3}{f(3)-f(x)}}\\&=x\left({\frac {f(3)-f(x)}{x-3}}\right)^{-1}\\&=x\left({\frac {(-1)(f(x)-f(3))}{x-3}}\right)^{-1}\\&=x(-1)^{-1}\left({\frac {f(x)-f(3)}{x-3}}\right)^{-1}\\&=-x\left({\frac {f(x)-f(3)}{x-3}}\right)^{-1}\end{aligned}}$ We now recognize the second term as the difference quotient. Hence, applying the limit laws, we arrive at our final answer ${\begin{aligned}\lim _{x\to 3}{\frac {x^{2}-3x}{f(3)-f(x)}}&=\lim _{x\to 3}\left[-x\left({\frac {f(x)-f(3)}{x-3}}\right)^{-1}\right]\\&=-\left(\lim _{x\to 3}x\right)\left(\lim _{x\to 3}{\frac {f(x)-f(3)}{x-3}}\right)^{-1}\\&=-3\left(f'(3)\right)^{-1}\\&=-3\left(5\right)^{-1}\\&=-{\frac {3}{5}}\end{aligned}}$

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