No worries, just use the homework section next time.
It's beena little while, but:
[tex]0^\infty[/tex]
and
[tex]\infty^0[/tex]
are also places that you can apply l'Hospitals rule.
Let's say we have two functions:
[tex]\lim_{x \rightarrow y} f(x) \rightarrow 0[/tex]
and
[tex]\lim_{x \rightarrow y} g(x) \rightarrow \infty[/tex]
Then
[tex]\lim_{x \rightarrow y} g(x)^{f(x)}=\lim_{x \rightarrow y} e^{f(x) \ln(g(x))}[/tex]
And the exponent there is of the form:
[tex]0 \times \infty[/tex]
The first thing that jumps into my head is that the limit, if you remove the 5^n, is simply 7.8^(n*2/n) = 60.84. So, since we've ignored adding 5^n before rooting, the answer must be at least 60.84.
There's only one answer that satisfies that, clearly demonstrating why I love multiple choice tests so much.
n's going to infinity right? Easy way is piecewise. 7*8^n >> 5^n as n => infinity, so that term can be neglected (basically same is dividing out the 7*8^n, just slightly quicker and less rigorous). You get 7^(2/n)*(8^n)^(2/n) => 8^2 = 64.