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Limit of the function (1+7/x)^(x/3)

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            x
            -
            3
     /    7\ 
 lim |1 + -| 
x->oo\    x/

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

Limit((1 + 7/x)^(x/3), x, oo, dir='-')

Detail solution

Let's take the limit

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

transform
do replacement

u = \frac{x}{7}

then

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

=
=

\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{\frac{7 u}{3}}

\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{\frac{7 u}{3}}

\left(\left(\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{\frac{7}{3}}

The limit

\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}

is second remarkable limit, is equal to e ~ 2.718281828459045
then

\left(\left(\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{\frac{7}{3}} = e^{\frac{7}{3}}

The final answer:

\lim_{x \to \infty} \left(1 + \frac{7}{x}\right)^{\frac{x}{3}} = e^{\frac{7}{3}}

Lopital's rule

There is no sense to apply Lopital's rule to this function since there is no indeterminateness of 0/0 or oo/oo type

The graph

Plot the graph

Other limits x→0, -oo, +oo, 1

\lim_{x \to \infty} \left(1 + \frac{7}{x}\right)^{\frac{x}{3}} = e^{\frac{7}{3}}

\lim_{x \to 0^-} \left(1 + \frac{7}{x}\right)^{\frac{x}{3}} = 1

\lim_{x \to 0^+} \left(1 + \frac{7}{x}\right)^{\frac{x}{3}} = 1

\lim_{x \to 1^-} \left(1 + \frac{7}{x}\right)^{\frac{x}{3}} = 2

\lim_{x \to 1^+} \left(1 + \frac{7}{x}\right)^{\frac{x}{3}} = 2

\lim_{x \to -\infty} \left(1 + \frac{7}{x}\right)^{\frac{x}{3}} = e^{\frac{7}{3}}

Rapid solution [src]

 7/3
e

e^{\frac{7}{3}}

Expand and simplify

The graph