How can I simplify this limit to infinity problem?

[tmt1]

I have this problem

$$\lim_{{x}\to{\infty}} \frac{2x + 1}{ \sqrt{x^2 + 2x + 1} + x}$$

How do I get from there to this step?

$$\lim_{{x}\to{\infty}} \frac{2x}{ \sqrt{x^2 } + x}$$

From the last step I can calculate them limit as 1.

 

[Prove It]

I have this problem

$$\lim_{{x}\to{\infty}} \frac{2x + 1}{ \sqrt{x^2 + 2x + 1} + x}$$

How do I get from there to this step?

$$\lim_{{x}\to{\infty}} \frac{2x}{ \sqrt{x^2 } + x}$$

You CAN'T! These things are not equal!

First notice that since we are approaching infinity, x is clearly positive, so we can treat $\displaystyle \begin{align*} \sqrt{x^2} = x \end{align*}$.

$\displaystyle \begin{align*} \frac{2x + 1}{\sqrt{ x^2 + 2x + 1 } + x } &= \frac{\frac{2x+1}{x}}{\frac{\sqrt{x^2 + 2x + 1} + x}{x}} \\ &= \frac{2 + \frac{1}{x}}{\frac{\sqrt{x^2 + 2x + 1}}{\sqrt{x^2}} + \frac{x}{x} } \\ &= \frac{ 2 + \frac{1}{x}}{ \sqrt{ 1 + \frac{2}{x} + \frac{1}{x^2}} + 1 } \end{align*}$

Now what happens as $\displaystyle \begin{align*} x \to \infty \end{align*}$?

 

[tmt1]

You CAN'T! These things are not equal!

First notice that since we are approaching infinity, x is clearly positive, so we can treat $\displaystyle \begin{align*} \sqrt{x^2} = x \end{align*}$.

$\displaystyle \begin{align*} \frac{2x + 1}{\sqrt{ x^2 + 2x + 1 } + x } &= \frac{\frac{2x+1}{x}}{\frac{\sqrt{x^2 + 2x + 1} + x}{x}} \\ &= \frac{2 + \frac{1}{x}}{\frac{\sqrt{x^2 + 2x + 1}}{\sqrt{x^2}} + \frac{x}{x} } \\ &= \frac{ 2 + \frac{1}{x}}{ \sqrt{ 1 + \frac{2}{x} + \frac{1}{x^2}} + 1 } \end{align*}$

Now what happens as $\displaystyle \begin{align*} x \to \infty \end{align*}$?

I understand this problem even less than I thought I did so now I am a bit unprepared. Is it:

\(\displaystyle \frac{2}{\sqrt{1} + 1}\)

 

[Jameson]

What Prove It is doing is showing you how you have to manipulate expressions in order to find limits. It looks weird at first but there is a reason to doing so.

For example look at this to start with: \(\displaystyle \frac{2x+1}{\sqrt{x^2+2x+1}+x}\)

Just for the heck of it, let's pull out an $x^2$ from the terms under the square-root.

\(\displaystyle \sqrt{x^2+2x+1}=\sqrt{x^2 \left(1+\frac{2}{x}+\frac{1}{x^2} \right)}\)

Why do this you might ask? Well it comes down to something similar to the last thread you made. Those two fractions we now have are going to drop off to $0$ as $x \rightarrow \infty$, leaving us with just $\sqrt{x^2}$. This is a technique you are going to have to use when working with limits at infinity to try to cancel some things and simplify some terms.

Does that help some?

 

[earboth]

I have this problem

$$\lim_{{x}\to{\infty}} \frac{2x + 1}{ \sqrt{x^2 + 2x + 1} + x}$$

...

Good morning,

your problem gives me the opportunity to show off - thanks:

\(\displaystyle \lim_{{x}\to{\infty}} \frac{2x + 1}{ \sqrt{x^2 + 2x + 1} + x} = \lim_{{x}\to{\infty}} \frac{2x + 1}{ \sqrt{(x+1)^2} + x} = \lim_{{x}\to{\infty}} \frac{2x + 1}{ x+1 + x} = \lim_{{x}\to{\infty}}( 1)\)

But be careful: This is not the general way to calculate limits!