limit of $\lim_{x\to 7}(\frac{x}{7})^{(\frac{x^2-18x+80}{x-7})}$ $$\lim_{x\to 7}\left(\frac{x}{7}\right)^{\large\left(\frac{x^2-18x+80}{x-7}\right)}$$
It is $1^{\infty}$
$$\lim_{x\to 7}\left(\frac{x}{7}\right)^{\large\left(\frac{(x-8)(x-10)}{x-7}\right)}$$
I tried to take 
$$\lim_{x\to 7}e^{\ln\left(\frac{x}{7}\right)^{\large\left(\frac{(x-8)(x-10)}{x-7}\right)}}=\lim_{x\to 7}e^{\large\left(\frac{(x-8)(x-10)}{x-7}\right)\ln\left(\frac{x}{7}\right)}$$
Now it is $e^{(0\cdot \infty)}$ which we can not conclude about the limit
 A: Hint:
\begin{align}
\lim_{x \to 7} (x-8)(x-10)\frac{\ln x - \ln 7}{x-7} = 3 \lim_{x \to 7} \frac{\ln x - \ln 7}{x-7}
\end{align}
A: $$\lim_{x\to 7}e^{\ln\left(\frac{x}{7}\right)^{\left(\frac{(x-8)(x-10)}{x-7}\right)}}=\lim_{x\to 7}e^{\left(\frac{(x-8)(x-10)}{x-7}\right)\ln\left(\frac{x}{7}\right)}$$
From your work we have:
$$\lim_{x\to 7}\left(\frac{(x-8)(x-10)}{x-7}\right)\ln\left(\frac{x}{7}\right)=\lim_{x\to 7}\left(\frac{(x-8)(x-10)}{1}\right)\frac{\ln\left(1+\color{red}{\left(\frac{x}{7}-1\right)}\right)}{\color{red}{\left(\frac{x}{7}-1\right)}}\cdot\frac{\left(x-7\right)}{7\left(x-7\right)}={\frac{3}{7}}$$
So the limit would be $\color{red}{\exp\left(\frac{3}{7}\right)}$
A: L'hopital
$\lim_{x\to 7}{(\frac{(x-8)(x-10)\ln(\frac{x}{7})}{x-7})}=$
$\lim_{x\to 7}\frac{(x-8)(x-10)\frac 7x\frac 17 + (2x-18)\ln \frac x7}1=$
$\frac{(-1)(-3)}7 + (-4)*0=\frac 37$.
So $\lim_{x\to 7}e^{ln(\frac{x}{7})^{(\frac{(x-8)(x-10)}{x-7})}}=\lim_{x\to 7}e^{(\frac{(x-8)(x-10)}{x-7})ln(\frac{x}{7})}=e^{\frac 37}$
That is.... assuming you did everything right.  I confess I didn't check  your work at all.
A: $\frac{(x-8)(x-10)}{x-7}\ \frac{\ln(1+\frac{x}{7}-1)}{\frac{x}{7}-1}\times (\frac{x}{7}-1)$
as $\frac{x}{7}\to 1\Rightarrow\displaystyle\lim_{x\rightarrow7} \frac{\ln(1+\frac{x}{7}-1)}{\frac{x}{7}-1}=1$
So we are left with $\frac{(x-8)(x-10)}{7(x-7)}\times(x-7)$ can you do from here.
If you have to find limit of $1^{\infty}$ which is $f(x)^{g(x)}$ , you can always remember it would $e^{g(x)\times(f(x)-1)}$
