## anonymous 3 years ago I know this limit should be zero but i'm not sure why. $\lim_{n \rightarrow \infty} \frac{ 917^(n+100) }{ 955^n }$

1. anonymous

Just to clarify that is 917^(n+100) on the top

2. anonymous

Just going to throw a joke out here because it's 4:04am... "The limit does not exist!" - Mean Girls

3. anonymous

That's not really the answer though :P

4. ash2326

$\frac{917^{n+100}}{955^{n}$ $n-> \infty$ ${\frac{917}{955}}^n \times 917^100$ $n\to \infty$ so ${\frac{917}{955} }^n->0$ $0\times 917^100=0$

5. ash2326

oops $0\times 917^{100}=0$

6. anonymous

Think i've got it, how i worked it out was like this. Could be completely wrong but oh well. $\frac{ n(\frac{ 917^{n+100} }{ n }) }{ 955 }$ $\frac{ 0 }{ 955 }= 0$ ?

7. anonymous

wait, left something out.

8. anonymous

ahhh, just imagine the bottom line as n(955)

9. ash2326

But this part $\frac{917^{n+100}}{n}\ne 0$ It's of the form $\frac{\infty}{\infty}$