# Limits at Infinity

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• Oct 14th 2009, 01:19 AM
mj.alawami
Limits at Infinity
Evaluate the following limits:
1)$\displaystyle lim (x\rightarrow \infty) \frac{\sqrt{x^2+4}}{x+4}$
2)$\displaystyle lim (x\rightarrow \infty) \frac{x^3-10}{x^2}$

Attempt:
For the first answer i rationalized the nummerator and got the answer as 1 (Wondering)
For the second answer I don't know how to fatorize it and go the denominato as zero and then got stuck

Thank you..
• Oct 14th 2009, 02:09 AM
CaptainBlack
Quote:

Originally Posted by mj.alawami
Evaluate the following limits:
1)$\displaystyle lim (x\rightarrow \infty) \frac{\sqrt{x^2+4}}{x+4}$
2)$\displaystyle lim (x\rightarrow \infty) \frac{x^3-10}{x^2}$

Attempt:
For the first answer i rationalized the nummerator and got the answer as 1 (Wondering)
For the second answer I don't know how to fatorize it and go the denominato as zero and then got stuck

Thank you..

For your first question divide top and bottom by $\displaystyle x$ to get:

$\displaystyle \lim_{x\rightarrow \infty} \frac{\sqrt{1+4/x^2}}{1+4/x}$

and now as both top and bottom go to a finite valuse as $\displaystyle x$ goes to infinity you should be able to find the limit.

For the second divide top and bottom by $\displaystyle x^2$, to get:

$\displaystyle \lim_{x\rightarrow \infty}\frac{ x-\frac{10}{x^2}}{1}$

and you should be able to see what happens now.

CB
• Oct 14th 2009, 02:42 AM
mj.alawami
Quote:

Originally Posted by CaptainBlack
For your first question divide top and bottom by $\displaystyle x$ to get:

$\displaystyle \lim_{x\rightarrow \infty} \frac{\sqrt{1+4/x^2}}{1+4/x}$

and now as both top and bottom go to a finite valuse as $\displaystyle x$ goes to infinity you should be able to find the limit.

For the second divide top and bottom by $\displaystyle x^2$, to get:

$\displaystyle \lim_{x\rightarrow \infty}\frac{ x-\frac{10}{x^2}}{1}$

and you should be able to see what happens now.

CB

But i was taught that i should choose the highest polynomial and divide it like for the first one i should divide by X and the second x^3
• Oct 14th 2009, 03:33 AM
Plato
Quote:

Originally Posted by mj.alawami
But i was taught that i should choose the highest polynomial and divide it like for the first one i should divide by X and the second x^3

But are these polynomials? No, the numerator is not.
• Oct 14th 2009, 03:33 AM
mj.alawami
Quote:

Originally Posted by CaptainBlack
For your first question divide top and bottom by $\displaystyle x$ to get:

$\displaystyle \lim_{x\rightarrow \infty} \frac{\sqrt{1+4/x^2}}{1+4/x}$

and now as both top and bottom go to a finite valuse as $\displaystyle x$ goes to infinity you should be able to find the limit.

For the second divide top and bottom by $\displaystyle x^2$, to get:

$\displaystyle \lim_{x\rightarrow \infty}\frac{ x-\frac{10}{x^2}}{1}$

and you should be able to see what happens now.

CB

So both answers will be 0
• Oct 14th 2009, 03:39 AM
Plato
Quote:

Originally Posted by mj.alawami
So both answers will be 0

No again. The limit is 1 in the first one.
• Oct 14th 2009, 03:47 AM
mr fantastic
Quote:

Originally Posted by mj.alawami
So both answers will be 0

Quote:

Originally Posted by Plato
No again. The limit is 1 in the first one.

@OP: And the limit of the second should be perfectly plain (substitute some large values of x - what happens?) Unfortunately you're allowing preconceived ideas based on cookbook recipes to muddy the waters.
• Oct 14th 2009, 04:14 AM
mj.alawami
Quote:

Originally Posted by Plato
No again. The limit is 1 in the first one.

• Oct 14th 2009, 04:18 AM
The Second Solution
Quote:

Originally Posted by mj.alawami
Please can you show me how did you do both. (Show your steps the gave you the final answer) because i got confused

In the first you're expected to know that 4/x and 4/x^2 --> 0 as x --> oo.

In the second, it has been suggested that you substitute some large values of x and see what happens. It doesn't look like you took that advice. If you had, the limit would become obvious.
• Oct 14th 2009, 04:21 AM
Plato
In $\displaystyle C$ is a constant and $\displaystyle N$ is a positive integer then
$\displaystyle \lim _{x \to \infty } \frac{C}{{x^N }} = 0$.

That is the crux of #1.
$\displaystyle \lim _{x \to \infty } \frac{4}{{x^2 }} = 0$

$\displaystyle \lim _{x \to \infty } \frac{4}{{x }} = 0$
• Oct 14th 2009, 04:24 AM
mj.alawami
Quote:

Originally Posted by CaptainBlack
For your first question divide top and bottom by $\displaystyle x$ to get:

$\displaystyle \lim_{x\rightarrow \infty} \frac{\sqrt{1+4/x^2}}{1+4/x}$

and now as both top and bottom go to a finite valuse as $\displaystyle x$ goes to infinity you should be able to find the limit.

For the second divide top and bottom by $\displaystyle x^2$, to get:

$\displaystyle \lim_{x\rightarrow \infty}\frac{ x-\frac{10}{x^2}}{1}$

and you should be able to see what happens now.

CB

1) This cancels to 1/sqrt(x + 4) which gets smaller and smaller as x increases so the limit is 0.

2) You have a rational function with the top polynomial of higher order than the bottom one. This means that the top will increase faster than the bottom as x increases so there is no limit. (Some might say a limit of infinity but I don't like that.)
• Oct 14th 2009, 04:28 AM
mj.alawami
Quote:

Originally Posted by Plato
In $\displaystyle C$ is a constant and $\displaystyle N$ is a positive integer then
$\displaystyle \lim _{x \to \infty } \frac{C}{{x^N }} = 0$.

That is the crux of #1.
$\displaystyle \lim _{x \to \infty } \frac{4}{{x^2 }} = 0$

$\displaystyle \lim _{x \to \infty } \frac{4}{{x }} = 0$

$\displaystyle \frac{\sqrt{1+4/x^2}}{1/x+4/x^2}$<<< I don't know what to do after this point
• Oct 14th 2009, 04:37 AM
Defunkt
Quote:

Originally Posted by mj.alawami

$\displaystyle \frac{\sqrt{1+4/x^2}}{1/x+4/x^2}$<<< I don't know what to do after this point

$\displaystyle \frac{\sqrt{x^2+4}}{x+4} = \frac{\sqrt{x^2+4}\cdot \frac{1}{x}}{(x+4) \cdot \frac{1}{x}}$ $\displaystyle = \frac{\sqrt{1+\frac{4}{x^2}}}{1+\frac{4}{x}}$

There is no need to further simplify.

$\displaystyle \lim_{x \to +\infty} \sqrt{1+\frac{4}{x^2}} = 1$, since $\displaystyle \lim_{x \to +\infty} \frac{4}{x^2}=0$.

Do the same for the denominator and you end up with $\displaystyle \frac{1}{1}$.
• Oct 14th 2009, 04:43 AM
mj.alawami
Quote:

Originally Posted by Defunkt
$\displaystyle \frac{\sqrt{x^2+4}}{x+4} = \frac{\sqrt{x^2+4}\cdot \frac{1}{x}}{(x+4) \cdot \frac{1}{x}}$ $\displaystyle = \frac{\sqrt{1+\frac{4}{x^2}}}{1+\frac{4}{x}}$

There is no need to further simplify.

$\displaystyle \lim_{x \to +\infty} \sqrt{1+\frac{4}{x^2}} = 1$, since $\displaystyle \lim_{x \to +\infty} \frac{4}{x^2}=0$.

Do the same for the denominator and you end up with $\displaystyle \frac{1}{1}$.

thanks
So the second answer is also 1
• Oct 14th 2009, 05:41 AM
Defunkt
Quote:

Originally Posted by mj.alawami
thanks
So the second answer is also 1

No

$\displaystyle \lim_{x\to +\infty} x - \frac{10}{x^2} = \lim_{x\to +\infty} x - \lim_{x\to +\infty} \frac{10}{x^2}$

Now, what is $\displaystyle \lim_{x\to +\infty} x$ ?

And what is $\displaystyle \lim_{x\to +\infty} \frac{10}{x^2}$ ?
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