Limit of (x+a / x+ 2a)^x when x approaches infinity

[Jane1728]

Hello,

I've encountered this question in my calculus book, the image of the resolution is attached to this post.
I did not understand very clearly steps (2) and (3) (they are marked in the end of each equation).

I know that: the limit of (1 + 1/x)^x when x approaches infinity is equal to e. So, what I understood from this question is that I have to transform the equation until I can apply that rule, but I do not understand how X^x appeared, and how it developed into step (3).

Thank you.

 

[topsquark]

They are flipping a fraction over. It's a bit weird looking but they are trying to get the expression into a particular form.

[math]\dfrac{a}{x} = \dfrac{ 1 }{ \left ( \dfrac{x}{a} \right )}[/math]
-Dan

 

[pka]

I've encountered this question in my calculus book, the image of the resolution is attached to this post.I did not understand very clearly steps (2) and (3) (they are marked in the end of each equation).

Jane, you have made far too complicated.
\(\dfrac{x+a}{x+2a}=1+\dfrac{-a}{x+2a}\) then all we need is to know that:
\(\Large{\mathop {\lim }\limits_{x \to \infty } {\left( {1 + \frac{c}{{x + d}}} \right)^{bx}} = {e^{bc}}}\)

 

[Jane1728]

They are flipping a fraction over. It's a bit weird looking but they are trying to get the expression into a particular form.

[math]\dfrac{a}{x} = \dfrac{ 1 }{ \left ( \dfrac{x}{a} \right )}[/math]
-Dan

Thank you for the reply, I understand it better now.

 

[Jane1728]

Jane, you have made far too complicated.
\(\dfrac{x+a}{x+2a}=1+\dfrac{-a}{x+2a}\) then all we need is to know that:
\(\Large{\mathop {\lim }\limits_{x \to \infty } {\left( {1 + \frac{c}{{x + d}}} \right)^{bx}} = {e^{bc}}}\)

That way is much easier! Thank you so much!

 

[Steven G]

At line 2 we can cancel out x^xx^xas it equals 1.

Now it is known that [math]\lim_{x\to 0}(1+\dfrac{a}{x})^x = e^a[/math]
Right from step 2 you have [math]\dfrac{e^a}{e^{2a}}...[/math]