# Thread: How to solve lim to infinity

1. ## How to solve lim to infinity

How to solve : $$\lim_{x \to \infty} (1 + 1/5x)^{2x + 6}$$

I know that
$$\lim_{x \to \infty} (1 + 1/x)^{x} = e$$
$$\lim_{x \to \infty} (1 + 1/x)^{3x} = e^3$$

I tried to differentiate it
$$(2x + 6) (1 + 1/5x)^{2x + 5}(1/5)$$

2. ## Re: How to solve lim to infinity

$$\lim_ {x\to\infty} (1 + a/x)^{x} = e^a$$

3. ## Re: How to solve lim to infinity

Originally Posted by Helly123
How to solve : $$\lim_{x \to \infty} (1 + 1/5x)^{2x + 6}$$

I know that
$$\lim_{x \to \infty} (1 + 1/x)^{x} = e$$
$$\lim_{x \to \infty} (1 + 1/x)^{3x} = e^3$$

I tried to differentiate it
$$(2x + 6) (1 + 1/5x)^{2x + 5}(1/5)$$
$u^{2x+6} = u^6\cdot \left(u^x\right)^2$

$\displaystyle \lim_{x \to \infty} u^{2x+6} = \lim_{x \to \infty} u^6\left( u^x \right)^2 = \left(\lim_{x \to \infty} u^6 \right) \left( \lim_{x \to \infty} u^x \right)^2$

4. ## Re: How to solve lim to infinity

So if you have $\displaystyle \lim_{x\to\infty}(1+\frac{1}{ax})^{bx+c}$, you substitute $\displaystyle y=ax$ to get:

$\displaystyle \lim_{x\to\infty}(1+\frac{1}{ax})^{bx+c}=\lim_{y \to \infty}(1+\frac{1}{y})^{\frac{b}{a}y+c}=((\lim_{y \to \infty}(1+\frac{1}{y})^y)^{\frac{b}{a}})(\lim_{y \to \infty}(1+\frac{1}{y})^c)=e^\frac{b}{a}$

since as $\displaystyle y$ goes to infinity, $\displaystyle (1+\frac{1}{y})^c$ goes to 1.

- Hollywood