A Fraction Has To Be Made

Calculus Level 3

We know that the limit of ratios of the two functions $\sqrt{x^2+x+1}$ and $\sqrt{x^2-x-1}$ as $x$ approaches infinity is equal to 1. That is,

$\displaystyle \lim_{x\to\infty} \dfrac{ \sqrt{x^2+x+1} }{\sqrt{x^2-x-1}} = \lim_{x\to\infty} \sqrt{ \dfrac{1 + 1/x + 1/x^2}{1 - 1/x - 1/x^2} } = 1 .$

Is it also true that the limit of the differences of the two same functions as $x$ approaches infinity is also equal to 1? That is, is the following limit true?

$\displaystyle \lim_{x\to\infty} \left ({ \sqrt{x^2+x+1} } - {\sqrt{x^2-x-1}} \right) = 1$

Yes, it is true No, it is not true

2 solutions

Aditya Narayan Sharma
Jun 25, 2016

Relevant wiki: Limits by Rationalization

Rationalising we get , $\displaystyle S=\lim_{x\to \infty} \frac{2(x+1)}{\sqrt{x^2+x+1}+\sqrt{x^2-x-1}}$

So dividing by $x$ up and down we get , $\displaystyle S=\lim_{x\to\infty} \frac{2(1+\frac{1}{x})}{\sqrt{1+\frac{1}{x}+\frac{1}{x^2}}+\sqrt{1-\frac{1}{x}-\frac{1}{x^2}}} = \frac{2}{2}=1$

Great! This is the easiest technique to solve this problem.

Challenge Master Note: Can you think of another technique?

Pi Han Goh - 4 years, 11 months ago

factor out $x^2 ,$ we get

$\quad L = \displaystyle \lim_{ x \to \infty} x\left( \left[ 1+\left( \dfrac{1}{x}+\dfrac{1}{x^2} \right) \right]^{\frac12}- \left[1+\left(-\dfrac{1}{x}-\dfrac{1}{x^2} \right) \right]^{\frac12}\right)$

Use binomial expansion: $L = \displaystyle \lim_{ x \to \infty} x \left( 1+\dfrac12\cdot\left( \dfrac{1}{x}+\dfrac{1}{x^2} \right)+\cdots-1-\dfrac{1}{2}\cdot\left(-\dfrac{1}{x}-\dfrac{1}{x^2} \right)-\cdots \right) \\ L =\displaystyle \lim_{ x \to \infty} x\left(\dfrac1{x}+\dfrac{1}{x^2}+\cdots \right) = \boxed{1}$

Sabhrant Sachan - 4 years, 11 months ago

Stolz-Cesaro theorem is another way to do this

Aditya Narayan Sharma - 4 years, 11 months ago

@Pi Han Goh This problem is useless, and this solution is useless because there are too many people who dont know the limit functions well in the world, and I belong to the people I said.

. . - 3 months, 1 week ago

So you want me to enable your request because you don't know something and you don't want to learn about it?

Pi Han Goh - 3 months, 1 week ago

X X
Apr 9, 2018

$\displaystyle \lim_{x\to\infty} \left ({ \sqrt{x^2+x+1}-(x+0.5) } \right) = \lim_{x\to\infty} \left ({ \sqrt{ (x+0.5)^{2}+0.75}-(x+0.5) } \right) = \lim_{x\to\infty} \left ( (x+0.5)-(x+0.5) \right)=0$ $\displaystyle \lim_{x\to\infty} \left ({ \sqrt{x^2-x-1}-(x-0.5) } \right) = \lim_{x\to\infty} \left ({ \sqrt{ (x-0.5)^{2}-1.25}-(x-0.5) } \right) = \lim_{x\to\infty} \left ( (x-0.5)-(x-0.5) \right)=0$ $\displaystyle \lim_{x\to\infty} \left ({ \sqrt{x^2+x+1} } - {\sqrt{x^2-x-1}} \right) =\lim_{x\to\infty} \left ({ [\sqrt{x^2+x+1}-(x+0.5)+(x+0.5)] } - {[\sqrt{x^2-x-1}-(x-0.5)+(x-0.5)]} \right)= \lim_{x\to\infty} \left ( (x+0.5)-(x-0.5) \right)=1$

Good attempt, but you just committed a fallacy.

$\displaystyle \lim_{x\to\infty} ( f(x) - g(x)) = \left (\lim_{x\to\infty} f(x)\right ) - \left (\lim_{x\to\infty} g(x) \right)$ is only true if both $\displaystyle \lim_{x\to\infty} f(x)$ and $\displaystyle \lim_{x\to\infty} g(x)$ are finite.

Pi Han Goh - 3 years, 1 month ago

I made a little correction to my solution.

X X - 3 years, 1 month ago

Great! It's fixed!!! Thanksss +1

Pi Han Goh - 3 years, 1 month ago

A Fraction Has To Be Made

2 solutions

0 pending reports