Consider the function defined by

Determine whether the sequence converges or diverges, and if it converges find the limit.

We use the definition of the exponential and some tricks to get into a form we know how to deal with,

The final line follows since

since

Therefore,

Thus, converges to the limit 0.

Would it still be valid to express the question as

and just claim the limits of both the numerator and denominator approach 0?

And? it’s an indeterminate form so your only hope is L’Hopital’s rule which makes it worse.

Here is my solution: if a is zero the limit is trivial, so suppose a>0, then the limit becomes n/e^(-nloga) and since abs(a)0 hence by theorem 7.11 the limit is zero. Now if a0 so the limit of n(-a)^n is zero by the argument above so for every epsilon>0 there is a delta>0 such that if n>delta then abs(n(-a)^n)<epsilon this proves this case since abs(n(a)^n)=abs(n(-a)^n)