Find the limit of 1/(x^2 - 9) as x tends to -3 from the left side.
Approaching -3 from the left means that the values of x must be slightly less than -3.
I created a table for x and f(x).
x...............(-4.5)..........(-4)...............(-3.5)
f(x).......... 0.088..........0.142....….....0.3076
I can see that f(x) is getting larger and larger and possibly without bound.
I say the limit is positive infinity.
Yes?
llaffer
Favorite Answer
Yes, but you'd want to test with numbers very close to -3. Like -3.001 :
1 / (x² - 9)
1 / ((-3.001)² - 9)
1 / (9.006001 - 9)
1 / 0.006001
166.63889
And do this again with -3.00001 (even closer to -3), if this goes to positive infinity we should have a much larger number:
1 / (x² - 9)
1 / ((-3.00001)² - 9)
1 / (9.0000600001 - 9)
1 / 0.0000600001
16666.63889
And it is, this shows that the limit would go to positive infinity.
husoski
Yes, and when x is slightly less than -3 (for example when x=-3.1) then |x| > 3 and x^2 > 9. Subtract 9 and you find that x^2 - 9 > 0, and of course 1/(x^2 - 9) > 0 also. The nearby values to the left are all positive so the "extended real" limit is +oo.