+84 Determine (if it exists) by evaluating the corresponding one-sided limits.
lim | x-3 |
x --> 3 ---------------
x^2 -9
a. limit does not exist
b. 0
c. 3
d. 1/3
e. -1/6
complete the table and use the result to estimate the equation numerically
lim x+2
x--> -2 ---------------
x^2 - 2x - 8
x -2.1 -2.01 -2.001 -2 -1.999 -1.99 -1.9
f(x) ?
a. infinity
b. -1/6
c. limit does not exist
d. -6
e. 6
+118690 Determine (if it exists) by evaluating the corresponding one-sided limits.
lim | x-3 |
x --> 3 ---------------
x^2 -9
As x tends to 3 from above x-3 is positive so |x-3|=x-3
so
\(\displaystyle\lim_{x\rightarrow3^+}\;\frac{|x-3|}{x^2-9}\\ =\displaystyle\lim_{x\rightarrow3^+}\;\frac{x-3}{x^2-9}\\ =\displaystyle\lim_{x\rightarrow3^+}\;\frac{1}{2x}\\ =\frac{1}{6}\)
BUT as x tends to 3 from below x-3 is negative so |x-3|=3-x
\(\displaystyle\lim_{x\rightarrow3^-}\;\frac{|x-3|}{x^2-9}\\ =\displaystyle\lim_{x\rightarrow3^-}\;\frac{3-x}{x^2-9}\\ =\displaystyle\lim_{x\rightarrow3^-}\;\frac{-1}{2x}\\ =\frac{-1}{6}\)
This means that the graph is not continuous at x=3 so the limit does not exist.
Here is a picture of the graph.
