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

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MATH 180 December 2017

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• December 2017 •

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 $\lim _{x\to 0}{\frac {f(2+3x)+f(2+5x)}{x}}$ .

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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
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: ${\textstyle \lim \limits _{x\to 0}{\dfrac {f(2+3x)+f(2+5x)}{x}}={\color {blue}32}}$ .