# Thread: Limit as X aproaches - infinity

1. ## Limit as X aproaches - infinity

I had this problem on a test, and it was only an alteration of a problem i had on a homework. But for some reason, if i use the same method to figure it, i get the wrong answer. I know the answer is 2/3 but i don't know the steps to solve it.

Limit 2+10^x
x→-∞ 3-10^x

= 2/3

My problem is how, >.< i can't figure it out. If it was positive infinity i would multiply by 1 or 10^-x/10^-x and get -1, but that doesn't seem to work. What am i missing?

2. Originally Posted by mortalapeman
I had this problem on a test, and it was only an alteration of a problem i had on a homework. But for some reason, if i use the same method to figure it, i get the wrong answer. I know the answer is 2/3 but i don't know the steps to solve it.

Limit 2+10^x
x→-∞ 3-10^x

= 2/3

My problem is how, >.< i can't figure it out. If it was positive infinity i would multiply by 1 or 10^-x/10^-x and get -1, but that doesn't seem to work. What am i missing?
$\displaystyle \lim_{x \rightarrow -\infty} 10^x = 0 \, ......$

3. Originally Posted by mr fantastic
$\displaystyle \lim_{x \rightarrow -\infty} 10^x = 0 \, ......$
well i hadn't thought about it that way....and that kinda just makes sense now >.<

My teacher kinda sucks hard and only repeats the examples in the book. So thank you for showing me.