Find $\lim _{x \to \infty}\frac{5^x+7^{x+1}}{7^x+5^{x+1}}$ 
I just started studying and in our first maths-exam we had to find the following limit:

$\lim \limits_{x \to \infty}\frac{5^x+7^{x+1}}{7^x+5^{x+1}}$

The other tasks where rather easy, so I feel like I'm missing something obvious here, but I don't even really know how to start with this one.
A walkthrough would be great, but hints in the right direction would also help.
Thank you!
 A: HINT
$$\frac{5^x+7^{x+1}}{7^x+5^{x+1}}=\frac{7^{x+1}}{7^x}\frac{\frac17\left(\frac57\right)^x+1}{1+5\left(\frac57\right)^x}=7\frac{\frac17\left(\frac57\right)^x+1}{1+5\left(\frac57\right)^x}$$
A: $\lim \limits_{x \to \infty}\frac{5^x+7^{x+1}}{7^x+5^{x+1}}$=$\lim \limits_{x \to \infty}\frac{(5/7)^x(1/7)+1}{1/7+(5/7)^{x+1}}=\frac{0+1}{1/7+0}=7$
A: Notice that:
$$\lim_{x\to\infty}\frac{5^x+7^{x+1}}{7^x+5^{x+1}}=\lim_{x\to\infty}\frac{(\frac{5}{7})^x+7}{1+\frac{5^{x+1}}{7^x}}=\frac{\lim_{x\to\infty}(\frac{5}{7})^x+7}{\lim_{x\to\infty}1+\frac{5^{x+1}}{7^x}}$$
This is done by dividing both the numerator and denominator by $7^x$. Can you finish the problem?
A: I would divide both numerator and denominator by $7^{x+1}$:
\begin{align*}
\frac{5^x+7^{x+1}}{7^x+5^{x+1}} &= \frac{\frac{5^x}{7^{x+1}} + \frac{7^{x+1}}{7^{x+1}}}{\frac{7^x}{7^{x+1}} + \frac{5^{x+1}}{7^{x+1}}}\\
&= \frac{\frac{1}{7}\left(\frac{5}{7}\right)^x + 1}{\frac{1}{7} + \left(\frac{5}{7}\right)^{x+1}}
\end{align*}
Since $\frac{5}{7}<1$, you know what $\left(\frac{5}{7}\right)^x$ tend to as $x\to\infty$.
A: When $x$ is large the dominant term in num. is $z^{x+1}$ and in den,it is $7^x$ so the limit $L=7$.
A: You can evaluate the limit as follows
$$\lim_{x\to \infty}\frac{5^x+7^{x+1}}{7^x+5^{x+1}}=\lim_{x\to \infty}\dfrac{7^x\left(\left(\frac{5}{7}\right)^x+7\right)}{7^x\left(1+5\left(\frac{5}{7}\right)^x\right)}$$
$$=\lim_{x\to \infty}\dfrac{7^x\left(0+7\right)}{7^x(1+5(0))}$$
$$=\lim_{x\to \infty}\left(\dfrac{7}{7}\right)^x7$$
$$=(1)7$$
$$=\color{blue}{7}$$
