Question

lim e^-x + 5/4^1^x as x approaches infinity

Answer 1

\[ \lim_{x\rightarrow -3} \frac{x^2}{x^2-9}\]

If you try:
\[ \lim_{x\rightarrow -3^-} \frac{x^2}{x^2-9}=\infty\]

\[ \lim_{x\rightarrow -3^+} \frac{x^2}{x^2-9} =- \infty\]

What does that say about the 2-sides limit ?

Answer 2

If you put x = -3, the numerator is +9 but the denominator is 0. Therefore, the limit is infinity. But you have to see if the limit is +infinity or -infinity by approaching -3 from both sides. If the limits are different then the limit does not exist.

Answer 3

it is approaching -3 from the right

Answer 4

but i did not know how to put that in the question

Answer 5

If you approach -3 from the left, you can start with x = -4 and approach -3. When x = -4, the numerator and denominator are both positive and so the limit when we approach x = -3 from the left is +infinity (which is what the first reply mentions in the middle section).

When we approach -3 from the right, we can start with x = -2 which makes the numerator positive but denominator negative. Therefore, when we approach x = -3 from the right, the limit is -infinity.

Since the left and right limits are different, the limit as x approaches -3 does not exist.

Answer 6

thanks