Science:Math Exam Resources/Courses/MATH104/December 2011/Question 01 (a)

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MATH104 December 2011

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Question 01 (a)
Compute $\lim _{x\rightarrow 4}{\frac {x^{2}-3x-4}{{\sqrt {x}}-2}}$

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
How can we use conjugates to deal with the square root in the bottom?

Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution 1

The limit evaluates to the form 0/0, ${\begin{aligned}\lim _{x\rightarrow 4}{\frac {x^{2}-3x-4}{{\sqrt {x}}-2}}&={\frac {4^{2}-3(4)-4}{{\sqrt {4}}-2}}\\&={\frac {16-12-4}{2-2}}\\&={\frac {0}{0}}\end{aligned}}$ which is undefined. However, notice that in the denominator we can get rid of the square root by multiplying top and bottom by the conjugate $\lim _{x\rightarrow 4}{\frac {x^{2}-3x-4}{{\sqrt {x}}-2}}\cdot {\frac {{\sqrt {x}}+2}{{\sqrt {x}}+2}}=\lim _{x\rightarrow 4}{\frac {(x^{2}-3x-4)({\sqrt {x}}+2)}{x-4}}.$ Now notice that the quadratic polynomial on the top can factor, $\displaystyle {}x^{2}-3x-4=(x-4)(x+1)$ and therefore ${\begin{aligned}&\lim _{x\rightarrow 4}{\frac {(x^{2}-3x-4)({\sqrt {x}}+2)}{x-4}}\\&=\lim _{x\rightarrow 4}{\frac {(x-4)(x+1))({\sqrt {x}}+2)}{x-4}}\\&=\lim _{x\rightarrow 4}(x+1)({\sqrt {x}}+2)\\&=20.\end{aligned}}$ Therefore $\lim _{x\rightarrow 4}{\frac {x^{2}-3x-4}{{\sqrt {x}}-2}}={\frac {4^{2}-3(4)-4}{{\sqrt {4}}-2}}=20.$

Solution 2

The limit evaluates to the form 0/0, ${\begin{aligned}\lim _{x\rightarrow 4}{\frac {x^{2}-3x-4}{{\sqrt {x}}-2}}&={\frac {4^{2}-3(4)-4}{{\sqrt {4}}-2}}\\&={\frac {16-12-4}{2-2}}\\&={\frac {0}{0}}\end{aligned}}$ which is undefined. 0/0 is a form for which we can apply l'Hospital's rule (if you know it): ${\begin{aligned}\lim _{x\rightarrow 4}{\frac {x^{2}-3x-4}{{\sqrt {x}}-2}}&=\lim _{x\rightarrow 4}{\frac {{\frac {d}{dx}}{\Big (}x^{2}-3x-4{\Big )}}{{\frac {d}{dx}}{\Big (}{\sqrt {x}}-2{\Big )}}}\\&=\lim _{x\rightarrow 4}{\frac {2x-3}{{\frac {1}{2}}x^{-1/2}}}\\&={\frac {5}{{\frac {1}{2}}4^{-1/2}}}\\&={\frac {5}{1/4}}\\&=20\end{aligned}}$

Solution 3

We can be a bit clever and factor $x^{2}-3x-4=(x-4)(x+1)$ then factor $x-4$ as a difference of squares where we think of $x$ as the square of ${\sqrt {x}}$. Hence $x-4=({\sqrt {x}}-2)({\sqrt {x}}+2)$. Combining gives ${\begin{aligned}\lim _{x\rightarrow 4}{\frac {x^{2}-3x-4}{{\sqrt {x}}-2}}&=\lim _{x\rightarrow 4}{\frac {(x-4)(x+1)}{{\sqrt {x}}-2}}\\&=\lim _{x\rightarrow 4}{\frac {({\sqrt {x}}-2)({\sqrt {x}}+2)(x+1)}{{\sqrt {x}}-2}}\\&=\lim _{x\rightarrow 4}({\sqrt {x}}+2)(x+1)\\&=({\sqrt {4}}+2)(4+1)\\&=20\end{aligned}}$

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