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Thread: How to solve lim to infinity

  1. #1
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    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
    $$(2x + 6) (1 + 1/5x)^{2x + 5}(1/5)$$
    Last edited by Helly123; Mar 7th 2018 at 09:39 PM.
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  2. #2
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    Re: How to solve lim to infinity

    $$\lim_ {x\to\infty} (1 + a/x)^{x} = e^a$$
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  3. #3
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    Re: How to solve lim to infinity

    Quote Originally Posted by Helly123 View Post
    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
    $$(2x + 6) (1 + 1/5x)^{2x + 5}(1/5)$$
    $u^{2x+6} = u^6\cdot \left(u^x\right)^2$

    $\displaystyle \lim_{x \to \infty} u^{2x+6} = \lim_{x \to \infty} u^6\left( u^x \right)^2 = \left(\lim_{x \to \infty} u^6 \right) \left( \lim_{x \to \infty} u^x \right)^2$
    Thanks from Helly123
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  4. #4
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    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
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