Log function Big-Oh

I need to show that $\log(1+5\epsilon) = O(\epsilon)$ as $\epsilon \rightarrow 0$ from the right hand side.

I believe I can do this using Taylor's theorem or L'Hopitals rule.

If the original statement $\log(1+5\epsilon) = O(\epsilon)$ as $\epsilon \rightarrow 0^+$ is true, then $\lim_{\epsilon \rightarrow 0^+}$$\frac{\log(1+5\epsilon)}{\epsilon}$ must be finite.

I have $\lim_{\epsilon \rightarrow 0^+}$$\frac{\log(1+5\epsilon)}{\epsilon}$ = $\lim_{\epsilon \rightarrow 0^+}$$\frac{5\epsilon - \frac{5}{2} \epsilon^2 + \frac{5}{3} \epsilon^3+.....}{\epsilon} = \lim_{\epsilon \rightarrow 0^+} 5 - \frac{5}{2} \epsilon + \frac{5}{3} \epsilon^2+.... = 5 < \infty $

So $\log(1+5\epsilon) = O(\epsilon)$ as $\epsilon \rightarrow 0^+$ must be true. Is this correct?

Another way I thought of solving was to write: $\log(1+5\epsilon) = 5\epsilon - \frac{5}{2} \epsilon^2 + \frac{5}{3} \epsilon^3+..... \approx 5\epsilon$ as terms of order $2$ and above become negligible as $\epsilon \rightarrow 0^+$. Could I then say that $|\log (1+5\epsilon)| \approx |5\epsilon| = 5 |\epsilon| \leq K |\epsilon|$ where $K= \frac{1}{5} \text{and} 0<|x|<\delta$ making this true?