Evaluating $\lim_{t \to 0} \frac{e^{5t} -1} {t}$

Is there a way to evaluate the following limit without using L'Hopital's rule? $$\lim_{t \to 0} \frac{e^{5t} -1} {t}$$

Based on this answer here I'm guessing that the series expansion for $e^{5t}$ would be helpful, but I'm not sure how to properly use it (I don't quite understand the rationale for the way the inequalities are setup).

Aside from series expansion, are there any other methods?

If you use $f(x)=e^{5x}$ then $f'(x)=5e^{5x}$

$$\lim_{t \to 0} \frac{e^{5t} - e^{5 \cdot 0}} {t-0} = \lim_{t \to 0} \frac{f(t) - f(0)} {t-0}=f'(0)=5e^{5\cdot 0}=5$$

Note that, $$\lim_{t\to 0}\dfrac{e^{5t} - 1}{t} = \lim_{5t\to 0}\dfrac{5(e^{5t} - 1)}{5t}$$ Here you can apply direct result that, $\lim_{x\to0} \dfrac{e^x-1}{x} = 1$ $$\implies 5(1) = 5$$

You also can use the series expansion $e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}=1+x+\frac{x^2}{2!}+\cdot\cdot\cdot$, so we have $$e^{5t}=\sum_{n=0}^{\infty}\frac{(5t)^n}{n!}=1+5t+\frac{(5t)^2}{2!}+\cdot\cdot\cdot$$

Thus $$\frac{e^{5t}-1}{t}=5+\frac{5^2}{2!}t+\cdot\cdot\cdot$$

Now take the limit.