Evaluating $\lim_\limits{x\to 0}(e^{5/x}-6x)^{x/2}$. Is my method correct? I have to solve the following limit which is in indeterminate form. \begin{equation}\lim_\limits{x\to 0}(e^{\frac{5}x}-6x)^{\frac{x}{2}} \end{equation}
So to solve this limit I attempted to do the following, I ended up making use of a substitution, however, I am not sure I did it right... Please let me know what you think (by the way my TI-nspire CX CAS already gave me the solution of $e^{\frac52}$.)
Work
\begin{align} L &=e^{\lim_\limits{x\to 0^+}(\frac{1}2 x\ln(e^{\frac{5}x-6x}))} \\ u&=\frac{1}x \\ L&=e^{\lim_\limits{u\to+\infty} \frac{1}{2}\frac{\ln(e^{5u}-6\frac{1}u)}{u}} \end{align}
From here I then did the following:
\begin{align} L&=e^{\lim_\limits{u\to+\infty} \frac12 \frac{5\frac{e^{5u}-\frac{6}{u^2}}{e^{5u}-\frac{6}{u}}}1}\end{align}
From Here I multiplied and divided by $\frac{\frac{1}{e^{5u}}}{\frac{1}{e^{5u}}}$
Leading me to this:
\begin{equation}L=e^{\frac{5}{2}\lim_\limits{u\to+\infty}\frac{1-\frac{6}{u^2e^{5u}}}{1-\frac{6}{ue^{5u}}}} \end{equation}
From Here I deduced that the answer was $L=e^{\frac{5}2}$, were my steps wrong, is infinity times infinity indeterminate and it requires further calculation please let me know.

You are correct, another way of writing this would be by using equivalences, since $6x=\underset{0}{o}\left( e^{\frac{5}{x}}\right)$, you get :
$$ e^{\frac{5}{x}}-6x \underset{x\rightarrow0}{\sim}e^{\frac{5}{x}}. $$ So you get :
$$(e^{\frac{5}{x}}-6x)^{\frac{x}{2}} \underset{x\rightarrow0}{\sim} e^{\frac{5}{2}},$$
so :
$$\lim_{x\rightarrow0}(e^{\frac{5}{x}}-6x)^{\frac{x}{2}} = e^{\frac{5}{2}}.$$
