limit involving logarithmic function

[mxrie]

hello, so i was trying to answer this
but i don’t understand how i’m supposed to evaluate the log part. i know i have to substitute -2 to x, giving me a log_1/3(0) but how do i find a limit for this?

thanks in advance

 

[Steven G]

You actually get log_1/3(0⁺) not log_1/3(0)
So what does the log_1/3(x) equal when x is small but positive?
Do you understand why I say 0⁺?

 

[mxrie]

You actually get log_1/3(0⁺) not log_1/3(0)
So what does the log_1/3(x) equal when x is small but positive?
Do you understand why I say 0⁺?

it's 0⁺ because x approaching -2 from the left would mean that it's square is greater than 4 right? would the limit be positive infinity (basing on the graph)?

i'm just a little confused because i thought the infinite limit would only happen if x approaches 0 or infinity

 

[HallsofIvy]

i'm just a little confused because i thought the infinite limit would only happen if x approaches 0 or infinity

That is certainly NOT true! Consider \(\displaystyle \lim_{x\to 2}\frac{1}{x- 2}\).

 

[Steven G]

it's 0⁺ because x approaching -2 from the left would mean that it's square is greater than 4 right? would the limit be positive infinity (basing on the graph)?

i'm just a little confused because i thought the infinite limit would only happen if x approaches 0 or infinity

Yes and then no. No, the limit would not be positive infinity based on the graph. Look again at the graph and the answer should be obvious.