What is the Limit of a Rational Expression?

Click For Summary

Discussion Overview

The discussion revolves around evaluating the limit of a rational expression as \( x \) approaches infinity, specifically the expression \( \frac{\sqrt{3x^2 - 1}}{x-1} \). Participants explore different methods and steps to arrive at the limit, which they suggest is \( \sqrt{3} \). The conversation includes technical reasoning and various approaches to the problem.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Exploratory

Main Points Raised

  • Some participants propose that the limit can be evaluated directly as \( \lim_{{x}\to{\infty}}\frac{\sqrt{3x^2 - 1 }}{x-1}=\sqrt{3} \).
  • Others detail a more complex approach involving dividing by \( x^2 \) and manipulating the expression to simplify the limit calculation.
  • A participant mentions the common technique of multiplying by the square root in both the numerator and denominator when dealing with square roots.
  • One participant expresses appreciation for the detailed steps provided by another, indicating that they learned from the process.
  • Another participant notes that they reached a similar step in their reasoning but did not fully understand the transition to the limit.

Areas of Agreement / Disagreement

Participants generally agree on the limit being \( \sqrt{3} \), but there are multiple approaches discussed without a consensus on the best method to arrive at that conclusion. The discussion remains exploratory with various techniques presented.

Contextual Notes

Some participants leave out absolute values in their calculations, assuming positivity as \( x \) approaches infinity. There are also indications of differing levels of understanding regarding the steps involved in the limit evaluation.

karush
Gold Member
MHB
Messages
3,240
Reaction score
5
$$\lim_{{x}\to{\infty}}\frac{\sqrt{3x^2 - 1 }}{x-1}=\sqrt{3}$$

I tried dividing everything by ${x}^{2}$ but not
 
Physics news on Phys.org
karush said:
$$\lim_{{x}\to{\infty}}\frac{\sqrt{3x^2 - 1 }}{x-1}=\sqrt{3}$$

I tried dividing everything by ${x}^{2}$ but not

$$\lim_{x \to +\infty} \frac{\sqrt{3x^2-1}}{x-1}= \lim_{x \to +\infty} \frac{3x^2-1}{(x-1) \sqrt{3x^2-1}}= \lim_{x \to +\infty} \frac{x^2 \left( 3- \frac{1}{x^2}\right)}{(x-1) \sqrt{x^2 \left( 3-\frac{1}{x^2}\right)}}= \lim_{x \to +\infty} \frac{x^2 \left( 3- \frac{1}{x^2}\right)}{(x-1)|x| \sqrt{ \left( 3-\frac{1}{x^2}\right)}}\\= \lim_{x \to +\infty} \frac{x^2 \left( 3- \frac{1}{x^2}\right)}{(x-1)x \sqrt{ \left( 3-\frac{1}{x^2}\right)}}= \lim_{x \to +\infty} \frac{x^2}{x^2 \left(1- \frac{1}{x} \right)} \cdot \lim_{x \to +\infty} \frac{3-\frac{1}{x^2}}{\sqrt{3-\frac{1}{x^2}}}= 1\cdot \frac{3}{\sqrt{3}}= \sqrt{3}$$
 
Wow, thanks, that was a lot of steps.
How would you know initially to go in that direction.
 
karush said:
Wow, thanks, that was a lot of steps.

(Smile)

karush said:
How would you know initially to go in that direction.
Usually when you have a square root you multiply by it at the numerator and the denominator.
 
Could you move this thread to the correct category?
 
karush said:
$$\lim_{{x}\to{\infty}}\frac{\sqrt{3x^2 - 1 }}{x-1}=\sqrt{3}$$

I tried dividing everything by ${x}^{2}$ but not

The easier way (first note that I'm leaving out absolute values because everything is positive when going to infinity anyway...)

$\displaystyle \begin{align*} \frac{\sqrt{3x^2 - 1}}{x - 1} &= \frac{\frac{1}{x}\,\sqrt{3x^2 - 1}}{\frac{1}{x} \left( x - 1 \right) } \\ &= \frac{\frac{1}{\sqrt{x^2}}\,\sqrt{3x^2 - 1}}{1 - \frac{1}{x}} \\ &= \frac{\sqrt{\frac{3x^2 - 1}{x^2}}}{1 - \frac{1}{x}} \\ &= \frac{\sqrt{3 - \frac{1}{x^2}}}{1 - \frac{1}{x}} \end{align*}$

So as $\displaystyle \begin{align*} x \to \infty , \, \frac{1}{x} \to 0 \end{align*}$ and $\displaystyle \begin{align*} \frac{1}{x^2} \to 0 \end{align*}$, giving $\displaystyle \begin{align*} \frac{\sqrt{3 - 0}}{1 - 0} = \sqrt{3} \end{align*}$ as the limit :)
 
I actually got to your 3rd step before posted, just didn't see the magic.

I sure learn a lot with MHB
 

Similar threads

High School Wondering if a limit exists or not
  • · Replies 3 ·
Replies
3
Views
1K
Undergrad Different Substitution for Solving an Integral
  • · Replies 6 ·
Replies
6
Views
3K
Undergrad How Do You Integrate Expressions Like These?
  • · Replies 6 ·
Replies
6
Views
2K
High School Rationale Behind t-Substitution for Evaluating Limits?
  • · Replies 5 ·
Replies
5
Views
2K
Undergrad Calculate limits as distributions
  • · Replies 4 ·
Replies
4
Views
2K
Graduate Solving a power series
  • · Replies 3 ·
Replies
3
Views
4K
Undergrad Question about Limit: \lim_{x\rightarrow1} (x+1)
  • · Replies 29 ·
Replies
29
Views
3K
Undergrad Express the limit as a definite integral
  • · Replies 16 ·
Replies
16
Views
4K
MHB Can this project revolutionize surfing?
  • · Replies 8 ·
Replies
8
Views
1K
Undergrad Finding the limit of a rational expression
  • · Replies 2 ·
Replies
2
Views
1K