Quiz 4 (Solution)


Date: MATH 1210-4 - Spring 2004

  1. $ \lim_{x\rightarrow 5} \frac{2x^2-11x+5}{x-5} = 9$. Give an $ \epsilon$-$ \delta$ proof of this fact.

    Given $ \epsilon>0$ we have to find $ \delta>0$ so that if $ 0< \vert x-5\vert
<\delta$, then $ \vert\frac{2x^2-11x+5}{x-5} -9\vert<\epsilon$.

    We have

    \begin{displaymath}\begin{split}\vert\frac{2x^2-11x+5}{x-5} -9\vert &= \vert\fra...
...2 \vert\frac{(x-5)^2}{x-5}\vert = 2 \vert x-5\vert. \end{split}\end{displaymath}    

    Now, $ \vert x-5\vert<\delta$, so that $ 2\vert x-5\vert<\epsilon$ if we take $ \delta
=\epsilon/2$.

    All together, for $ \delta
=\epsilon/2$, if $ 0<\vert x-5\vert<\delta
=\epsilon/2$ we have $ \vert\frac{2x^2-11x+5}{x-5} -9\vert = 2\vert x-5\vert <2
\epsilon/2 = \epsilon$.

  2. Knowing that $ \lim_{x\rightarrow 7} f(x) = 1$ and $ \lim_{x\rightarrow 7} g(x) = 9$, use properties of the limit operation to find

    $\displaystyle \lim_{x\rightarrow 7} \frac{2 (g(x)-4f(x))^2}{\sqrt{g(x)}}.$    

    Since the expression is a quotient, we compute the limit of the quotient as the quotient of the limits, provided that the limit in the denominator is $ \neq 0$ (this will become clear in the computation).

    \begin{displaymath}\begin{split}\lim_{x\rightarrow 7} \frac{2(g(x)-4f(x))^2}{\sq...
...-4\cdot 1)^2}{3} = \frac{2(5)^2}{3} = \frac{50}{3}. \end{split}\end{displaymath}    



Javier Fernandez 2004-02-13