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.
 A: Assuming $x \to 0^+$, your way is a little bit long and can be simplified but it is fine indeed at the end we obtain
$$\frac{1-\frac{6}{u^2 \ e^{5u}}}{1-\frac{6}{u \ e^{5u}}} \to 1$$
since $\frac{6}{u^2 \ e^{5u}}\to 0$ and $\frac{6}{u \ e^{5u}} \to 0$.
As an alternative, more directly we have
$$(e^{\frac{5}x}-6x)^{\frac{x}{2}}=(e^{\frac{5}x})^{\frac{x}{2}}\left(1-\frac{6x}{e^{\frac{5}x}}\right)^{\frac{x}{2}} =e^\frac52\left(1-\frac{6x}{e^{\frac{5}x}}\right)^{\frac{x}{2}} \to e^\frac52 \cdot 1 = e^\frac52$$
where it is important to note that the following
$$\lim_\limits{x\to 0^+}\left(1-\frac{6x}{e^{\frac{5}x}}\right)^{\frac{x}{2}}=1^0=1$$
is not an indeterminate form.

Note that for $x\to 0^-$ by $y=-x \to 0^+$ we obtain
$$\lim_\limits{x\to 0^-}(e^{\frac{5}x}-6x)^{\frac{x}{2}}=\lim_\limits{y\to 0^+}\frac1{\left(6y+\frac1{e^{\frac{5}y}}\right)^{\frac{y}{2}}}=1$$
since
$$\left(6y+\frac1{e^{\frac{5}y}}\right)^{\frac{y}{2}}=\left(6y\right)^{\frac{y}{2}}\left(1+\frac1{6ye^{\frac{5}y}}\right)^{\frac{y}{2}} \to 1\cdot(1+0)^0=1$$
A: 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 :
$$\lim_{x\rightarrow0}(e^{\frac{5}{x}}-6x)^{\frac{x}{2}} = e^{\frac{5}{2}}.$$
