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

Question

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

That's not really the answer though :P

ash2326 · Answer

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

ash2326 · Answer

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

anonymous · Answer

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

anonymous · Answer

wait, left something out.

anonymous · Answer

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

ash2326 · Answer

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