Limit of sequence is indeterminate.

Question

We have to find the limit of the sequence $a_k = \left(\dfrac{k^4 11^k + k^9 9^k}{7^{2k} +1}\right)$.

Here is my attempt:

$$\left(\frac{k^4 11^k + k^9 9^k}{7^{2k} +1}\right) = \frac{k^4 11^k + k^9 9^k}{49^k +1} = \frac{\frac{\left(\frac{11}{49}\right)^k}{k^5} + \left(\frac{9}{49}\right)^k}{\frac{1}{k^9} + \frac{1}{k^9 49^k}}$$

But as $k \to \infty $, both the numerator and the denominator tend to zero. So I am left with an indeterminate form. How can I avoid this?

Answer 1

Guide:

$$\left|\frac{k^4 11^k + k^9 9^k}{7^{2k} +1}\right| \le \frac{k^4 11^k + k^9 9^k}{49^k } = k^4 \left( \frac{11}{49}\right)^k + k^9 \left( \frac{9}{49}\right)^k$$

You can focus on finding $\lim_{k \to \infty}k^4 \left( \frac{11}{49}\right)^k$ and $\lim_{k \to \infty}k^9 \left( \frac{9}{49}\right)^k$

or in general $\lim_{k \to \infty} k^a r^k$ where $|r|<1$ and $a \ge 0$.

Answer 2

$$ \lim_ {k\to \infty}a_k =\lim_ {k\to \infty} \left(\frac{k^4 11^k + k^9 9^k}{7^{2k} +1}\right)=\lim_ {k\to \infty}\left(\frac{k^4 ( 11/{49})^k + k^9 (9/{49})^k}{1+7^{-2k} }\right) =0 $$