help with limits please

Question

experimentx · Answer

take 17^t common from inside

anonymous · Answer

ln?

experimentx · Answer

no ... take 17^t common on numerator.

experimentx · Answer

Woops!! hold on a sec .. your t->-inf

experimentx · Answer

according to wolf .. this diverges.

experimentx · Answer

showing this is enough ... http://www.wolframalpha.com/input/?i=Limit%5B%28%2810%5Et+%2B+17%5Et%29%29%5Et%2C+t+-%3E+-Infinity%5D gotta sleep ... 5 am here.

experimentx · Answer

woops!! sorry again ...i put the wrong expoentiation

anonymous · Answer

thank you! sorry i had to take out the food from the oven. ill look over this right now

experimentx · Answer

http://www.wolframalpha.com/input/?i=Limit%5B%28%2810%5Et+%2B+17%5Et%29%2F2%29%5E%281%2Ft%29%2C+t+-%3E+-Infinity%5D

anonymous · Answer

1/t right?

experimentx · Answer

Showing this much is enough ... since  
$$ 2^{- 1 /\infty} = 2^0 = 1$$

anonymous · Answer

oh okk i got it thanks :)

experimentx · Answer

to show that 
$$ \lim_{t\to-\infty } \left( 10^t + 17^t\right)^{1/t}$$

first factor out 10^t ... you get

$$ \lim_{t\to-\infty } 10 \left( 1 + (17/10)^t\right)^{1/t}$$
the inner vlaue (17/10)^t -> 0 as x->-inf
and 1/t = 0 as x->0 ... you get 1. multiply it by 10 outside, you will get 10 as answer.

BB GN ... gotta sleep 5 am here.

anonymous · Answer

got it, i thought it was positive infinity ! at first. ty and good night