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# limit when denominator is 0?

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Hello!

I'm having difficulty finding the limit of function x / ((x^2) - 4)) as x approaches 2 from the left (x--> 2-). My textbook tells me that the limit would be -infinite, but how does this work? With 0 in the denominator, wouldn't the limit not exist?

****Edit: I can see from graphing the function that x goes towards negative infinity as it approaches 2 from the left, but is there any way I could tell whether a limit is nonexistent/infinity without having to graph the function?

Thank you so much!

Jun 6, 2022
edited by Guest  Jun 6, 2022

#1
+118621
+1

Does this help?

$$\displaystyle \lim_{x\rightarrow2^-}\;\frac{x}{x^2-4}$$

If x is between 0 and 2 (not including 2)  then this will be  pos divided by neg  which is negative

that is why this limit HAS to be negative

$$\displaystyle \lim_{x\rightarrow2^-}\;\frac{x}{x^2-4}\\~\\ =​​\displaystyle \lim_{x\rightarrow2^-}\;\frac{x/x^2}{(x^2-4)/x^2}\\~\\ =​​\displaystyle \lim_{x\rightarrow2^-}\;\frac{1/x}{1-4/x^2}\\~\\ ==> \frac{approaching \;0.5}{appraoching\; 0\; from\; negative \;side}\\~\\ =-\infty$$

Jun 6, 2022
edited by Melody  Jun 6, 2022
#2
+118621
+1

Note:

You are NOT finding the value at x=2, you are finding the value as x approaches 2 (from below)

So it is not   0/0

and

infinity is not a number, it is a concept.

Jun 6, 2022
edited by Melody  Jun 6, 2022