Finding a limit with power: $ \lim_ {x \to \infty} \left( \frac {7x+10}{1+7x} \right)^{x/3} $

Question

i have tried dividing this limit by x but i do not know what to do next. Maybe you could help me?

$$ \lim_ {x \to \infty} \left( \frac {7x+10}{1+7x} \right)^{x/3} $$

Answer 1

• 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 $(1+\dfrac kx)^x\to e^k$

Answer 2

$\lim_{x\to\infty}\left( \frac {7x+10}{1+7x} \right)^{x/3}$

$\lim_{x\to\infty}\left( e^{\frac 13\cdot\ln(\frac {7x+10}{1+7x})^x} \right)$

$\implies\left( e^{\frac 13\cdot\lim_{x\to\infty}\ln\bigg(\frac {1+\frac{10}{7x}}{\frac1{7x}+1}\bigg)^x} \right)$

Use the fact that $\lim_{x\to\infty}\bigg(1+\frac1{ax}\bigg)^x=e^\frac1a$

$\implies\left( e^{\frac 13\cdot\lim_{x\to\infty}\ln\bigg(\frac {1+\frac{10}{7x}}{\frac1{7x}+1}\bigg)^x} \right) = \bigg(e^{\frac13\cdot\ln\bigg[\dfrac{e^\frac{10}7}{e^{\frac17}}\bigg]}\bigg)$

$= e^{\frac13\cdot\frac97} =e^\frac37$

Answer 3

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,$$\lim_{x\to\infty}\ln{y}=\lim_{x\to\infty}\frac{\frac{-63}{10 + 77 x + 49 x^2}}{-3/x^2}=\lim_{x\to\infty}\frac{63/3}{\frac{10}{x^2} + \frac{77}{x} + 49}=\frac{21}{49}=\frac{3}{7}$$ $$\ln\bigg(\lim_{x\to\infty}y\bigg)=\frac{3}{7}$$ $$\lim_{x\to\infty}y=e^{3/7}$$

Answer 4

$(1+\dfrac{9}{1+7x})^{x/3}$.

$y:= \dfrac{1+7x}{9}$, 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 $y \rightarrow \infty$.

Answer 5

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.

Answer 6

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}