Stuck on 4 Evaluate the Limit problems; Do you agree that all are DNE?

**Fenixx09** · August 28th 2011, 03:15 PM

I got DNE for all four of these (zero in the denominator). I am a little skeptical about submitting 4 answers as DNE, so I need your help.

I will show you what I have gotten done on these four. Can you let me know if they really are DNE, or if you were able to get further on certain problems (without telling me the answer).

[Problem 1]

$\\\begin{array}{c}\lim\\x\rightarrow-3\end{array} \ \ \frac{6}{x+3}$

Can't factor this one and don't see a reason to multiply by the conjugate, so I plugged in for X and got DNE.

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[Problem 2]

$\\\begin{array}{c}\lim\\x\rightarrow5\end{array} \ \ \frac{x-9}{x^{2}-14x+45} \\ \\ \\ \begin{array}{c}\lim\\x\rightarrow5\end{array} \ \ \frac{x-9}{(x-9)(x-5)} \\ \\ \\ \begin{array}{c}\lim\\x\rightarrow5\end{array} \ \ \frac{1}{(x-5)} \\ \\ \\ \frac{1}{5-5} \\ \\ \\ \frac{1}{0}$

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[Problem 3]

$\\\begin{array}{c}\lim\\x\rightarrow5\end{array} \ \ \frac{5x^{2}-125}{x^{2}-10x+25} \\ \\ \\ \begin{array}{c}\lim\\x\rightarrow5\end{array} \ \ \frac{5(x^{2}-25)}{(x-5)(x-5)} \\ \\ \\ \begin{array}{c}\lim\\x\rightarrow5\end{array} \ \ \frac{5(x-5)(x+5)}{(x-5)(x-5)} \\ \\ \\ \begin{array}{c}\lim\\x\rightarrow5\end{array} \ \ \frac{5(x+5)}{(x-5)} \\ \\ \\ \frac{50}{0}$

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[Problem 4]

$\\\begin{array}{c}\lim\\x\rightarrow4\end{array} \ \ \frac{x-2}{\sqrt{x}-2} \\ \\ \\ \begin{array}{c}\lim\\x\rightarrow4\end{array} \ \ \left(\frac{x-2}{\sqrt{x}-2}\right) \left(\frac{\sqrt{x}+2}{\sqrt{x}+2}\right) \\ \\ \\ \begin{array}{c}\lim\\x\rightarrow4\end{array} \ \ \frac{(x-2)(\sqrt{x}+2)}{x-4} \\ \\ \\ \frac{8}{0}$

**pickslides** · August 28th 2011, 03:31 PM

Think about the function $\displaystyle y =\frac{a}{x+b}$ , where is the asymptote?

**Fenixx09** · August 28th 2011, 03:42 PM

Originally Posted by pickslides

Think about the function $\displaystyle y =\frac{a}{x+b}$ , where is the asymptote?

Well, if you are referring to my first problem, the Vertical Asymptote is at x=-3. My teacher said that there are different techniques to finding a limit and that plugging the x value into the equation right away often gives "something"/"zero" and we don't want that because then the limit does not exist. As far as I can tell, it is best to get the "X" out of the denominator whenever I have a problem that would yield a zero in the denominator.

I'm not sure where you were headed by asking me about the asymptote, but thank you for responding. Really, I just wanted some of the "more learned" individuals on this forum to double check my work (because I really am doubting my ability to properly evaluate a limit when I get this many "DNE"s).

**TheChaz** · August 28th 2011, 05:13 PM

The reason for asking about the asymptote might be to get you to consider the graph of the function to direct and/or support your intuition.

The answer is "does not exist", because the denominator gets arbitrarily close to zero - both positive AND negative values. So one of hand (side) it goes up, and on the other hand (side) it goes down.

Hmm... after a quick glance, they all appear to have the same answer, for effectively the same reasoning.

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