Pre-Calc. Function. as x ->∞, 8^x approaches..? etc.?

Get your textbook out and read through the section on limits. The limits on the exam will likely require a bit more work than these questions. a. Yep, ∞. Think of 8^1,000,000,000, 8^1,000,000,000,000, ever bigger powers of 8 will just keep on going to ever bigger numbers. [Of course, I think the correct answer for the limit technically is "the limit does not exist."] Think of an exponential curve. b. Nope. Think of 8^(-1,000,000,000). That equals 1/(8^1,000,000,000). 1 divided by bigger and bigger and bigger numbers gets closer and closer and closer to 0. Again, think of an exponential curve. c. as x-> 0+ means x approaching 0 from the positive (greater than 0) side. Because 10^y equals a positive number for all positive and negative y, log(x) -- the inverse of 10^x -- would not be defined for negative numbers, so we can't talk about the limit as x->0-. [We are sticking to the Real numbers.] As x->0+, 8x->0+ also. The log of a very small number becomes a negativer and negativer number [Proper English -- who needs it? <grin>] Thank again about the exponential curve: you can get the y-value as close to 0 as you want as the x-value goes more and more towards -infinity. The log function is the inverse of exponentiation [or imagine mirroring the function across the line y=x]. d. The limit of (1/8)^x as x->∞ = limit (1/8^x)) as x->∞. So, when we divide 1 by bigger and bigger numbers, we get closer and closer to 0. This is also equal to limit of 8^(-x) as x->∞, which in turn is equal to the limit of 8^x as x->-∞, which was question b.

Your teacher's advice is good, but it's not the only way to do it. 8^x is > x for all large x. So it goes to infinity even faster than x does. That's how you answer a). When f goes to infinity, 1/f goes to 0. So 1/8^x goes to 0 as x goes to infinity. And by the way, as x goes to - infinity, 8^x goes to whatever 8^(-x) goes to when x goes to infinity. :) That shows the answers to b and d have to be the same, and also shows they're both 0.

Your teacher's advice is good, but it's not the only way to do it. 8^x is > x for all large x. So it goes to infinity even faster than x does. That's how you answer a). When f goes to infinity, 1/f goes to 0. So 1/8^x goes to 0 as x goes to infinity. And by the way, as x goes to - infinity, 8^x goes to whatever 8^(-x) goes to when x goes to infinity. :) That shows the answers to b and d have to be the same, and also shows they're both 0.