MHB How can I simplify this limit to infinity problem?

  • Thread starter Thread starter tmt1
  • Start date Start date
  • Tags Tags
    Infinity Limit
AI Thread Summary
The limit problem presented is $$\lim_{{x}\to{\infty}} \frac{2x + 1}{ \sqrt{x^2 + 2x + 1} + x}.$$To simplify, one can factor out \(x\) from both the numerator and the denominator, leading to $$\lim_{{x}\to{\infty}} \frac{2 + \frac{1}{x}}{\sqrt{1 + \frac{2}{x} + \frac{1}{x^2}} + 1}.$$As \(x\) approaches infinity, the limit simplifies to $$\frac{2}{\sqrt{1} + 1} = 1.$$The discussion emphasizes the importance of manipulating expressions correctly to find limits at infinity, particularly by recognizing dominant terms.
tmt1
Messages
230
Reaction score
0
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.
 
Mathematics news on Phys.org
tmt said:
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 $\displaystyle \begin{align*} x \to \infty \end{align*}$?
 
Prove It said:
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:

$$\frac{2}{\sqrt{1} + 1}$$
 
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: $$\frac{2x+1}{\sqrt{x^2+2x+1}+x}$$

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

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

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?
 
tmt said:
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:

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

But be careful: This is not the general way to calculate limits!
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Back
Top