1. ## Limits at Infinity

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

Thank you..

2. 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

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

3. 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

4. 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.

5. 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

6. Originally Posted by mj.alawami
So both answers will be 0
No again. The limit is 1 in the first one.

7. 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.

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

9. 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.

10. 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.
$\lim _{x \to \infty } \frac{4}{{x^2 }} = 0$

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

11. 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.)

12. 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.
$\lim _{x \to \infty } \frac{4}{{x^2 }} = 0$

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

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

13. Originally Posted by mj.alawami

$\frac{\sqrt{1+4/x^2}}{1/x+4/x^2}$<<< 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 $\frac{1}{1}$.

14. 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

15. 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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