anonymous 5 years ago lim x approaches infinity of (Square root of x^2 +2)/ x-1

1. anonymous

found a good video that helped with this problem http://www.youtube.com/watch?v=GCJxNkw7-Ec&feature=related

2. amistre64

lim x-->inf sqrt(x^2 +2)/x-1 Since x^2 +2 cannot be factored into "real" numbers it is a prime. Since nothing can be canceled out from top to bottom, there are no "holes" in the graph. sqrt(x^2 +2)/x-1 As x reaches infinity we get an indeterminite value: so I am wondering if we can apply L-Hopitals rule: lim x-->inf (f(x)/g(x)) = lim x-->inf (f'(x)/g'(x)).

3. anonymous

i just divided the top and bottoome byt x and got square root of x^2/x which is equal to absolute value of x divided by x and got 1 as the answer.

4. anonymous

*bottom

5. amistre64

That is the algebra way. The calculus way is also good to know :)

6. anonymous

what is it?

7. amistre64

It is called the L-Hopital rule and you find the derivative of the top and the derivative of the bottom. The derivative of the bottom (x-1) = 1 so that gets rid of that. But the top derives to (x/sqrt(x^2 +2)) which doesnt seem to help us out ... then again, I am still trying to learn it good :)

8. anonymous

that seems pretty simple. thanks for the explanation

9. anonymous

i have another problem that is simple, but i'm not sure how to solve it lim as x approaches -infinity of x^2+2/x-1

10. anonymous

and i've got to solve it algebraically

11. amistre64

algebraicaly you want to multiply this equation by "1" or rather some form of "1" that is useful. in order to do that; you take the highest power of "x" from the bottom and use it like this: (1/x)/(1/x) and this equals "1" When we multiply the top we get: (x^2)/x + (2/x) = x+(2/x). And the bottom becomes: (x/x) + (1/x) = 1+(1/x) As the value of "x" gets very big; the fractions with an "x" in the bottom get very very tiny so that they basically have no effect in the equations so we can throw them away. We are left with x/1. As x--> infinity; the limit is a diagonal line of (y=x). Some people call it an oblique asymptote or a slany asymptote.

12. amistre64

slant not "slany"... i got fat fingers lol

13. anonymous

lol, thanks alot u make much more sense than my professor

14. amistre64

thanx :) There is a general rule for finding the limits. Do you know them?

15. anonymous

nope

16. amistre64

lol...thats fine too

17. anonymous

my calc class is based on youtube and this website

18. amistre64

I thought I typed "want to know them" ... my mistake. Yeah, my calc class is based on what I have taught myself as well.

19. anonymous

no problem, sure, i'd like to know em'

20. amistre64

here they are: In the case that the highest exponent of x is in the bottom, for example: x+1/x^2 -2 When we multiply by the highest exponent of "x" from teh bottom we get: (1/x^2)/(1/x^2) the only numbers we care about here will be the stuff that does not end up with an x in the bottom of it. the top: (x+1)(1/x^2) = (x/x^2)+(1/x^2) = (1/x)+(1/x^2) all these have an "x" in the bottom of the fraction; so they all go to 0 right?. The bottom part: (x^2 -2)(1/x^2) = (x^2/x^2)-(2/x^2) = (1)-(2/x^2). The only number we have here that matters is the (1) because anything with an "x" in the bottom goes to "0". What we end up with is 0/1. So, when the highest exponent of "x" is in the bottom part of the original equation; the limit will always go to: 0/1 = 0

21. amistre64

If the highest exponent of "x" is the same on the top as it is on the bottom: everything goes away except for the coeeficients of the highest exponent of x. Example: (5x^2 +1)/(2x^2 -9x +12)... all we care about is (5x^2)/(2x^2) = 5/2 is your limit

22. anonymous

awesome, thanks alot. this will defintely help me out.

23. amistre64

the other case is this: the highest eponent of "x" is in the top part of the fraction. Example: (x^3 + (other things which dont matter)) / (x + (more stuff that doesnt matter). = (x^2)/x = x. Your limit as x--> infinity will equal the graph of y=x.

24. anonymous

for the second question, i asked you, wouldn't the limit just be infinity, since it equals x and x is approaching infinity.

25. amistre64

that should be (x^2 + (other things....))

26. amistre64

If we were only concerned about straight the horizontal asymptote, then YES. But, we can still predict what the graph will act like. And it acts like the graph for y=x; or some other curve. This type of curved limit is called an oblique asymptote. And it simply means that the limit is inifity but we know how the graph will act near infinity. Makes sense?

27. anonymous

somewhat, lol

28. anonymous

i think i got the jist of it though

29. amistre64

As you do more of them, you will do just fine :)

30. anonymous

thanks for all the help, i really appreciate it

31. anonymous

:-)