Science:Math Exam Resources/Courses/MATH110/December 2017/Question 02 (c)

MATH110 December 2017

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Question 02 (c)
Determine whether the following statements are true or false. If true, provide an explanation, if false, provide a counterexample. (c) True/False. If ${\textstyle f(x)=(x+1)^{100}}$ then ${\textstyle \displaystyle \lim _{h\to 0}{\frac {f(h)-f(0)}{h}}=100}$.

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
Recall the derivative of a function.

Hint 2
The limit coincides with the derivative of the function $f$ at the origin.

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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. By Hint 1, we recall that the derivative ${\textstyle f'(x)}$ of ${\textstyle f(x)}$ (at the point ${\textstyle x}$) is $$f'(x)=\lim _{h\to 0}{\dfrac {f(x+h)-f(x)}{h}}.$$ At the point ${\textstyle x=0}$, the derivative is $$f'(0)=\lim _{h\to 0}{\dfrac {f(h)-f(0)}{h}}.$$ Therefore, using the power rule, we have $$\lim _{h\to 0}{\dfrac {f(h)-f(0)}{h}}=f'(0)=\left.100(x+1)^{99}\right|_{x=0}=100(0+1)^{99}=100.$$ Answer: ${\textstyle {\color {blue}{\mbox{True}}}}$.