Science:Math Exam Resources/Courses/MATH 180/December 2017/Question 03 (c)

MATH 180 December 2017

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Question 03 (c)
(c) Let ${\textstyle f(x)}$ be such that ${\textstyle f(2)=0}$, ${\textstyle f'(2)=4}$ and ${\textstyle f'(x)}$ is continuous. Calculate $\lim _{x\to 0}{\frac {f(2+3x)+f(2+5x)}{x}}$.

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
Compute the limits of the numerator and denominator separately first.

Hint 2
Use L'Hôpital's rule.

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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 ${\textstyle g(x)=f(2+3x)+f(2+5x)}$ and ${\textstyle h(x)=x}$. First we compute the limits of ${\textstyle g(x)}$ and ${\textstyle h(x)}$ as ${\textstyle x\to 0}$: $$\lim \limits _{x\to 0}g(x)=\lim _{x\to 0}f(2+3x)+\lim _{x\to 0}f(2+5x)=f(2)+f(2)=0+0=0,$$ $$\lim _{x\to 0}h(x)=\lim _{x\to 0}x=0.$$ In the first equation, we have used the fact that since $f$ is differentiable (i.e.,${\textstyle f'(x)}$ exists for any $x$), ${\textstyle f}$ is continuous. We see that the desired limit is of the inderterminate form ${\textstyle {\frac {0}{0}}}$. Applying L’Hôpital’s rule, we have $$\lim \limits _{x\to 0}{\dfrac {g(x)}{h(x)}}=\lim \limits _{x\to 0}{\dfrac {g'(x)}{h'(x)}},$$ provided that the latter limit exists. Using chain rule with ${\textstyle g(x)=f(F(x))+f(G(x))}$, where ${\textstyle F(x)=2+3x}$, ${\textstyle G(x)=2+5x}$, ${\textstyle F'(x)=3}$ and ${\textstyle G'(x)=5}$, we have $$g'(x)=f'(F(x))F'(x)+f'(G(x))G'(x)=f'(2+3x)\cdot 3+f'(2+5x)\cdot 5.$$ On the other hand we observe that $$h'(x)=1.$$ At the limit ${\textstyle x\to 0}$, using the fact that ${\textstyle f'(x)}$ is continuous, we have $$\lim _{x\to 0}g'(x)=f'(2)\cdot 3+f'(2)\cdot 5=8f'(2)=8\cdot 4=32.$$ Also $$\lim _{x\to 0}h'(x)=1.$$ Therefore, $$\lim \limits _{x\to 0}{\dfrac {g(x)}{h(x)}}=\lim \limits _{x\to 0}{\dfrac {g'(x)}{h'(x)}}={\dfrac {32}{1}}=32.$$ Answer: ${\textstyle \lim \limits _{x\to 0}{\dfrac {f(2+3x)+f(2+5x)}{x}}={\color {blue}32}}$.