Limits at Infinity

**mj.alawami** · October 14th 2009, 01:19 AM

Evaluate the following limits:
1) $lim (x\rightarrow \infty) \frac{\sqrt{x^2+4}}{x+4}$
2) $lim (x\rightarrow \infty) \frac{x^3-10}{x^2}$

Attempt:
For the first answer i rationalized the nummerator and got the answer as 1

For the second answer I don't know how to fatorize it and go the denominato as zero and then got stuck

PLease show me how to answer these questions

Thank you..

**CaptainBlack** · October 14th 2009, 02:09 AM

Originally Posted by mj.alawami

Evaluate the following limits:
1) $lim (x\rightarrow \infty) \frac{\sqrt{x^2+4}}{x+4}$
2) $lim (x\rightarrow \infty) \frac{x^3-10}{x^2}$

Attempt:
For the first answer i rationalized the nummerator and got the answer as 1

For the second answer I don't know how to fatorize it and go the denominato as zero and then got stuck

PLease show me how to answer these questions

Thank you..

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

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

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

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

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

and you should be able to see what happens now.

CB

**mj.alawami** · October 14th 2009, 02:42 AM

Originally Posted by CaptainBlack

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

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

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

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

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

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

**Plato** · October 14th 2009, 03:33 AM

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.

**mj.alawami** · October 14th 2009, 03:33 AM

Originally Posted by CaptainBlack

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

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

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

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

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

and you should be able to see what happens now.

CB

So both answers will be 0

**Plato** · October 14th 2009, 03:39 AM

Originally Posted by mj.alawami

So both answers will be 0

No again. The limit is 1 in the first one.

**mr fantastic** · October 14th 2009, 03:47 AM

Originally Posted by mj.alawami

So both answers will be 0

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.

**mj.alawami** · October 14th 2009, 04:14 AM

Originally Posted by Plato

No again. The limit is 1 in the first one.

can you show me your simplifying method (in steps please)

**The Second Solution** · October 14th 2009, 04:18 AM

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.

**Plato** · October 14th 2009, 04:21 AM

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

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

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

**mj.alawami** · October 14th 2009, 04:24 AM

Originally Posted by CaptainBlack

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

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

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

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

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

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.)

**mj.alawami** · October 14th 2009, 04:28 AM

Originally Posted by Plato

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

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

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

Can you please show me how did you simplify your answer to get 1.

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

**Defunkt** · October 14th 2009, 04:37 AM

Originally Posted by mj.alawami

Can you please show me how did you simplify your answer to get 1.

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

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

There is no need to further simplify.

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

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

**mj.alawami** · October 14th 2009, 04:43 AM

Originally Posted by Defunkt

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

There is no need to further simplify.

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

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

thanks
So the second answer is also 1

**Defunkt** · October 14th 2009, 05:41 AM

Originally Posted by mj.alawami

thanks
So the second answer is also 1

No

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

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

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

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