Science:Math Exam Resources/Courses/MATH110/December 2015/Question 02 (b)

MATH110 December 2015

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Question 02 (b)
Evaluate the following limits or determine they are infinite or do not exist. Show all your work or provide justification. Answers without any accompanying work will receive no marks. (b) $\lim _{x\to 0}{\frac {3+x}{x-x^{2}}}$

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 hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
If the denominator of a fraction becomes zero while its numerator is nonzero, what does it tell us about its limiting behaviour?

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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. $\lim _{x\to 0}{\frac {3+x}{x-x^{2}}}={\frac {3}{0}}$ Since the denominator is getting smaller and smaller as $x\to 0$, the fraction is getting larger and larger, but we need to determine whether it get large positively or negatively. When $x$ approaches from the right, it is enough to consider $0<x<1$ and hence we have ${\frac {3+x}{x(1-x)}}>0$. This implies that $\lim _{x\to 0^{+}}{\frac {3+x}{x-x^{2}}}=\lim _{x\to 0^{+}}{\frac {3+x}{x(1-x)}}={\frac {3}{1\times 0^{+}}}=+\infty$ . On the other hand, when $x$ approaches from the left, we get $x<0$ and hence ${\frac {3+x}{x(1-x)}}<0$. This implies that $\lim _{x\to 0^{-}}{\frac {3+x}{x-x^{2}}}=\lim _{x\to 0^{-}}{\frac {3+x}{x(1-x)}}={\frac {3}{1\times 0^{-}}}=-\infty$ . Therefore, the answer is $\color {blue}DNE$.