Science:Math Exam Resources/Courses/MATH104/December 2015/Question 10 (a)

MATH104 December 2015

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Question 10 (a)
Let $f(x)=x^{2}+x-{\frac {1}{x}}-1$. Its derivative satisfies $f'(x)=2x+1+{\frac {1}{x^{2}}}={\frac {2x^{3}+x^{2}+1}{x^{2}}}={\frac {(x+1)(2x^{2}-x+1)}{x^{2}}}$. (a) Evaluate $\lim _{x\to 0^{-}}f(x)$ and $\lim _{x\to 0^{+}}f(x)$.

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
Consider the limits $\lim \nolimits _{x\to 0^{-}}x^{2}+x-1$ and $\lim \nolimits _{x\to 0^{-}}-{\tfrac {1}{x}}$ separately, and likewise for $\lim \nolimits _{x\to 0^{+}}.$

Hint 2
Notice that as $x\to 0^{-},$ the expression $-{\tfrac {1}{x}}$ assumes very large positive values. (Indeed, its value can be made arbitrarily large by choosing $x$ sufficiently close to (but still less than) 0.) What does this say about $\lim \nolimits _{x\to 0^{-}}-{\tfrac {1}{x}},$ and how can we adapt this reasoning to find $\lim \nolimits _{x\to 0^{+}}-{\tfrac {1}{x}}$ also?

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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. We observe that $\lim \nolimits _{x\to 0^{-}}-{\tfrac {1}{x}}=\infty$ and that $\lim \nolimits _{x\to 0^{+}}-{\tfrac {1}{x}}=-\infty .$ Hence ${\begin{aligned}\lim _{x\to 0^{-}}f(x)&=\lim _{x\to 0^{-}}x^{2}+x-1+\lim _{x\to 0^{-}}-{\frac {1}{x}}=-1+\infty ={\color {blue}\infty },\\\lim _{x\to 0^{+}}f(x)&=\lim _{x\to 0^{+}}x^{2}+x-1+\lim _{x\to 0^{+}}-{\frac {1}{x}}=-1+-\infty ={\color {blue}-\infty }.\end{aligned}}$