Finding a limit with power: $ \lim_ {x \to \infty} \left( \frac {7x+10}{1+7x} \right)^{x/3} $ i have tried dividing this limit by x but i do not know what to do next. Maybe you could help me?
$$ \lim_ {x \to \infty} \left( \frac {7x+10}{1+7x} \right)^{x/3}  $$
 A: *

*factor $7x$ in both numerator and denominator

*use expansion of $\dfrac 1{1+u}=1-u+o(u)$ to find $1+\dfrac ax+o(\frac 1x)$ inside parenthesis

*then use $(1+\dfrac kx)^x\to e^k$

A: One way to do: Let $y=(\frac{7x+10}{1+7x})^{x/3}$. Then $\ln{y}=\frac{x}{3}\ln{\frac{7x+10}{1+7x}}=\frac{\ln{\frac{7x+10}{7x+1}}}{3/x}$.
If you take the limit of $\ln{y}$, you will have $\frac{0}{0}$. By L'Hôpital's rule,$$\lim_{x\to\infty}\ln{y}=\lim_{x\to\infty}\frac{\frac{-63}{10 + 77 x + 49 x^2}}{-3/x^2}=\lim_{x\to\infty}\frac{63/3}{\frac{10}{x^2} + \frac{77}{x} + 49}=\frac{21}{49}=\frac{3}{7}$$ $$\ln\bigg(\lim_{x\to\infty}y\bigg)=\frac{3}{7}$$ $$\lim_{x\to\infty}y=e^{3/7}$$
A: $\lim_{x\to\infty}\left( \frac {7x+10}{1+7x} \right)^{x/3}$
$\lim_{x\to\infty}\left( e^{\frac 13\cdot\ln(\frac {7x+10}{1+7x})^x} \right)$
$\implies\left( e^{\frac 13\cdot\lim_{x\to\infty}\ln\bigg(\frac {1+\frac{10}{7x}}{\frac1{7x}+1}\bigg)^x} \right)$
Use the fact that $\lim_{x\to\infty}\bigg(1+\frac1{ax}\bigg)^x=e^\frac1a$
$\implies\left( e^{\frac 13\cdot\lim_{x\to\infty}\ln\bigg(\frac {1+\frac{10}{7x}}{\frac1{7x}+1}\bigg)^x} \right) = \bigg(e^{\frac13\cdot\ln\bigg[\dfrac{e^\frac{10}7}{e^{\frac17}}\bigg]}\bigg)$
$= e^{\frac13\cdot\frac97} =e^\frac37$
A: $(1+\dfrac{9}{1+7x})^{x/3}$.
$y:= \dfrac{1+7x}{9}$, then
$x /3= (9y -1)/21$, and
$(1+1/y)^{(y(9/21)-1)}=$
$[(1+1/y)^y]^{3/7} \cdot (1+1/y)^{-1}$.
Take the limit $y \rightarrow \infty$.
A: If you want to get more than the limit itself.
$$A= \left( \frac {10+7x}{1+7x} \right)^{x/3}=\left( 1+\frac {9}{1+7x} \right)^{x/3}$$ Take logarithms
$$\log(A)=\frac x 3 \log\left( 1+\frac {9}{1+7x} \right)$$ Use the Taylor expansion  $$\log(1+\epsilon)=\epsilon -\frac{\epsilon ^2}{2}+O\left(\epsilon ^3\right)$$ Make $\epsilon=\frac {9}{1+7x}$ and continue with long division to get
$$\log\left( 1+\frac {9}{1+7x} \right)=\frac{9}{7 x}-\frac{99}{98 x^2}+O\left(\frac{1}{x^3}\right)$$ $$\log(A)=\frac{3}{7}-\frac{33}{98 x}+O\left(\frac{1}{x^2}\right)$$ Continue with Taylor
$$A=e^{\log(A)}=e^{3/7}\left(1-\frac{33}{98 x}\right)+O\left(\frac{1}{x^2}\right)$$ which shows the limit and how it is approached.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
\lim_{x \to \infty}\pars{7x + 10 \over 1 + 7x}^{x/3} & =
\lim_{x \to \infty}\pars{21x + 10 \over 21x + 1}^{x} =
\lim_{x \to \infty}{\bracks{1 + \pars{10/21}/x}^{x} \over
\bracks{1 + \pars{1/21}/x}^{x}} = {\expo{10/21} \over \expo{1/21}}
\\[5mm] & =
\bbx{\expo{3/7}} \approx 1.5351
\end{align}
