Science:Math Exam Resources/Courses/MATH110/December 2014/Question 04 (b)

MATH110 December 2014

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Question 04 (b)
The following limit $$\lim _{h\to 0}{\frac {{\frac {1}{(2+h)^{2}}}-{\frac {1}{4}}}{h}}$$ equals the derivative ${\textstyle f'(a)}$ for some function ${\textstyle f}$ and some point ${\textstyle a}$. (b) Evaluate the limit above (without using any differentiation rules).

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?

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Hint
Simplify the numerator by using their common denominator.

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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, we simplify the denominator; ${\frac {1}{(2+h)^{2}}}-{\frac {1}{4}}={\frac {4-(2+h)^{2}}{4(2+h)^{2}}}={\frac {4-(h^{2}+4h+4)}{4(2+h)^{2}}}={\frac {-(h^{2}+4h))}{4(2+h)^{2}}}={\frac {-h(h+4)}{4(2+h)^{2}}}$ Now plug this back to the limit expression and then we obtain the limit as ${\begin{aligned}\lim _{h\to 0}{\frac {{\frac {1}{(2+h)^{2}}}-{\frac {1}{4}}}{h}}&=\lim _{h\to 0}{{\frac {1}{(2+h)^{2}}}-{\frac {1}{4}}}\cdot {\frac {1}{h}}\\&=\lim _{h\to 0}{\frac {-h(h+4)}{4(2+h)^{2}}}\cdot {\frac {1}{h}}=\lim _{h\to 0}{\frac {-(h+4)}{4(2+h)^{2}}}\cdot =-{\frac {4}{4\cdot 2^{2}}}=\color {blue}-{\frac {1}{4}}.\end{aligned}}$