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

MATH110 December 2015

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Question 02 (a)
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. (a) <math>\lim_{x\to3} \frac{x^2 -2x - 3}{x^2-9}

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
The first step of finding a limit is direct substitution. If substitution gives $0/0$, then we need to simplify by factoring, cancellation, etc and again substitute.

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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. Direct substitution gives $\lim _{x\to 3}{\frac {x^{2}-2x-3}{x^{2}-9}}={\frac {3^{2}-6-3}{3^{2}-9}}={\frac {0}{0}}$ However, both the numerator and denominator can be factored, so that we obtain ${\begin{aligned}\lim _{x\to 3}{\frac {x^{2}-2x-3}{x^{2}-9}}=\lim _{x\to 3}{\frac {(x-3)(x+1)}{(x-3)(x+3)}}=\lim _{x\to 3}{\frac {x+1}{x+3}}={\frac {4}{6}}=\color {blue}{\frac {2}{3}}\end{aligned}}$