How can I prove that for any $ \epsilon > 0$ exists: $\log_a(n) \le cn^\epsilon$?

Question

How can I prove that for any $ \epsilon > 0$ can I find $n_0 > 0$ , $c>0$ so that:

For any $n>n_0$ exists that:

$$\log_a(n) \le c\cdot n^\epsilon$$

NOTE: $a>1$.

Answer 1

$$\log_a(n) \le c\cdot n^\epsilon \iff \frac {\log_a(n)}{ n^\epsilon}<c$$

Apply L'Hospital Rule to get $$ \lim _{n\to \infty } \frac {\log_a(n)}{ n^\epsilon}=0 $$

Thus there exists an integer $N_0$ such that $$ n\ge N_0 \implies \frac {\log_a(n)}{ n^\epsilon}<c$$