## anonymous 5 years ago how can I delete indetermination in the following expression. Limx(x->0)((sqrt(x+2)-sqrt(2))/x)

1. anonymous

$limit _{x->0}(\sqrt{x+2}-\sqrt{2})/2$ this is the equation

2. anonymous

conjugation

3. anonymous

limitx−>0(x+2−−−−−√−2−−√)/x Sorry

4. anonymous

you mean by multiplyting the numerator and the denominator by $\sqrt{x+2}+\sqrt{2}$

5. anonymous

Top and bottom

6. anonymous

ok let me try

7. anonymous

Changuanas, I actually get rid of indetermination. However when I apply the limit x->0, the answer is not the corrrect. The answer int he book is 1/2sqrt(2).

8. anonymous

Sorry. I was wrong. I already got the answer! Can you help me with another exercise? Just to check

9. anonymous

just post question, if fall asleep someone else would help

10. anonymous

$Limit _{x->infinity} \sqrt{3x ^{4}-2x+4}+x$

11. anonymous

I don't remember what happens to infinity under square root. Does it go to infinity?

12. anonymous

According to the graph, it increases with no bound

13. anonymous

Ignore all others and consider only highest exponent of x. What is highest exp?

14. anonymous

I'm sorry. Did you already have an answer: infinity? or it needs more work?

15. anonymous

yes, more work :)

16. anonymous

I think the answer is 1

17. anonymous

what is highest exponent x in your problem?

18. anonymous

inside the root? its 2

19. anonymous

Ignore all others divide sq rt 3x^4 by sq rt x^4

20. anonymous

Wouldn't that one have to be just infinity?

21. anonymous

I think he is in an early class. The class before you study infinity, right Emun?

22. anonymous

yes

23. anonymous

In that case, don't ignore everything else, that is what you do in higher math;keep everything and divide each term by x^2

24. anonymous

It's funny being all done finals but still here during my break lol. why am I here?

25. anonymous

except that expression isn't anything you would need to do that for. maybe if you were dividing by a polynomial

26. anonymous

You here to help Emun I'm going to sleep.

27. anonymous

guys the equation was wrong. The equation is the following: $\sqrt{x ^{2}-2x+4}+x$ I factored the expression inside the root out and I got $\sqrt{(x-2)^{2}}+x$ so finally I got $x-2+x$ as I apply the limit as x=infinity,the answer I got is 2

28. anonymous

Thanks changaunas :)

29. anonymous

The answer would be infinity, wouldn't it?

30. anonymous

wouldn't you get 2x-2 , which as x becomes large.. your expression does too? infinity!

31. anonymous

Yeah and this is the answer: Thanks gusy

32. anonymous

thanks guys :)

33. anonymous

$\lim_{x \rightarrow 0} (\sqrt{x+2}-\sqrt{2})/x \times (\sqrt{x+2}+\sqrt{2})/(\sqrt{x+2}+\sqrt{2})$ $\lim_{x \rightarrow 0} (x+2-2)/(x(\sqrt{x+2}+\sqrt{2}))$ $\lim_{x \rightarrow 0} x/(x(\sqrt{x+2}+\sqrt{2}))$ $\lim_{x \rightarrow 0} 1/(\sqrt{x+2}+\sqrt{2})$ $1/(\sqrt{0+2}+\sqrt{2})$ $1/(\sqrt{2}+\sqrt{2})$ $1/(2\sqrt{2})$