# What is the limit of  ((1)+(3/x)+(5/x^2))^x as x approaches infinity?

Aug 7, 2016

${e}^{3}$

#### Explanation:

${\left(\left(1\right) + \left(\frac{3}{x}\right) + \left(\frac{5}{x} ^ 2\right)\right)}^{x} = {\left(1 + \frac{3 + \frac{5}{x}}{x}\right)}^{x}$

substituting $y = \frac{x}{3}$ we have

${\left(1 + \frac{3 + \frac{5}{x}}{x}\right)}^{x} \equiv {\left(1 + \frac{3 + \frac{5}{3 y}}{3 y}\right)}^{3 y} =$
${\left(1 + \frac{1 + \frac{5}{9 y}}{y}\right)}^{3 y} = {\left({\left(1 + \frac{1 + \frac{5}{9 y}}{y}\right)}^{y}\right)}^{3}$

then

${\lim}_{x \to \infty} {\left(1 + \frac{3 + \frac{5}{x}}{x}\right)}^{x} = {\lim}_{y \to \infty} {\left({\left(1 + \frac{1 + \frac{5}{9 y}}{y}\right)}^{y}\right)}^{3} =$
${\left({\lim}_{y \to \infty} {\left(1 + \frac{1 + \frac{5}{9 y}}{y}\right)}^{y}\right)}^{3} = {e}^{3}$

Note. We used the textbook result.

${\lim}_{y \to \infty} {\left(1 + \frac{1}{y}\right)}^{y} = e$

Aug 7, 2016

${e}^{3}$.

#### Explanation:

Reqd. Limit $= {\lim}_{x \rightarrow \infty} {\left(1 + \frac{3}{x} + \frac{5}{x} ^ 2\right)}^{x}$

$= {\lim}_{x \rightarrow \infty} {\left(1 + \frac{3 x + 5}{x} ^ 2\right)}^{x}$

$= {\lim}_{x \rightarrow \infty} {\left\{{\left(1 + \frac{3 x + 5}{x} ^ 2\right)}^{{x}^{2} / \left(3 x + 5\right)}\right\}}^{\frac{3 x + 5}{x}}$

=lim_(xrarroo){(1+(3x+5)/x^2)^(x^2/(3x+5))}^((3+5/x)

$= {\lim}_{x \rightarrow \infty} {\left\{{\left(1 + \frac{3 x + 5}{x} ^ 2\right)}^{{x}^{2} / \left(3 x + 5\right)}\right\}}^{3} \cdot {\left\{{\left(1 + \frac{3 x + 5}{x} ^ 2\right)}^{{x}^{2} / \left(3 x + 5\right)}\right\}}^{\frac{5}{x}}$

$= {e}^{3} \cdot {e}^{0}$

$= {e}^{3}$, as Cesareo R., Sir has derived!

Enjoy Maths!