$\lim_{x \rightarrow \infty}\frac{5^{x+1}+7^{x+1}}{5^x +7^x}$ $$\lim_{x \rightarrow \infty}\frac{5^{x+1}+7^{x+1}}{5^x +7^x}$$
I tried using L'Hospital rule,  which yielded :
$$\lim_{x \rightarrow \infty}\frac{5^{x+1} \ln 5+7^{x+1} \ln 7}{5^x \ln 5 }$$
But I'm at the dead end...  If I divide numerator and denominator by $5^x$ , I get a term $\frac{7^x}{5^x}$ ... which is unsolvable for the limit $x \rightarrow \infty $  
However , the answer provided by book is $-7$ . I doubt there is mistake in the question. 
 A: The standard method is to factor out and cancel the highest power, which is here $7^x$:
We get:
$\lim_{x\to\infty} \frac{5^{x+1}+7^{x+1}}{5^x+7^x}=\lim_{x\to\infty} \frac{7^{x+1}\left(\left(\frac57\right)^{x+1}+1\right)}{7^x\left(\left(\frac57\right)^x+1\right)}=\lim_{x\to\infty} \frac{7\left(\left(\frac57\right)^{x+1}+1\right)}{\left(\frac57\right)^x+1}$
$=7\cdot\lim_{x\to\infty} \frac{\left(\frac57\right)^{x+1}+1}{\left(\frac57\right)^x+1}$
It is $\lim_{x\to\infty} \left(\frac{5}{7}\right)^x=0$, since $\frac57<1$
We get:
$=7\cdot\frac{0+1}{0+1}=7$
A: $\dfrac{5^{x+1}+7^{x+1}}{5^{x}+7^{x}}=\dfrac{5(\frac{5}{7})^x+7}{(\frac{5}{7})^x+1}\to \dfrac{5\cdot0+7}{0+1}=7$
A: $$\lim_{x \rightarrow \infty}\frac{5^{x+1}+7^{x+1}}{5^x +7^x}=\lim_{x \rightarrow \infty}\frac{7^{x+1}({5^{x+1}\over 7^{x+1}}+1)}{7^{x}({5^x \over  7^{x}}+1)}$$
$$= 7\lim_{x \rightarrow \infty}\frac{{\big({5\over 7}\big)^{x+1}}+1}{\big({5\over 7}\big)^{x}+1}$$
$$= 7\lim_{t \rightarrow 0}\frac{{5\over 7}t+1}{t+1} =7$$
where $t= \big({5\over 7}\big)^{x}$
A: Simplify the fraction by $7^x$, and then everything will have a limit. 
The limit is $7$, not $-7$, so maybe there is a typo.
A: $$\lim_{x\to \infty}\frac{7^{x+1}((\frac57)^{x+1}+1)}{7^x((\frac57)^x+1)}$$
when $x\to \infty$ since $\frac57 \lt1$ then $(\frac57)^{x+1}\to 0$ similar with $(\frac57)^{x}$
Divide $7^{x+1}$ with $7^{x}$ which is 7, therefore, the limit is $7$
