## hpfan101 one year ago $\lim_{x \rightarrow \infty}(\sqrt{x^2-ax}-\sqrt{x^2+bx}$

1. jim_thompson5910

Hint: Multiply top and bottom by the conjugate $\large \lim_{x \rightarrow \infty}\left(\sqrt{x^2-ax}-\sqrt{x^2+bx}\right)$ $\large \lim_{x \rightarrow \infty}\left[\left(\sqrt{x^2-ax}-\sqrt{x^2+bx}\right) \color{red}{\times \frac{\sqrt{x^2-ax}+\sqrt{x^2+bx}}{\sqrt{x^2-ax}+\sqrt{x^2+bx}}}\right]$

2. anonymous

wow am i wrong!

3. hpfan101

Alright so i multiplied by the conjugate and when I simplified I got $\frac{ ax-bx }{ x \sqrt{ax}+x \sqrt{bx} }$

4. hpfan101

Apparently the answer is $\frac{ 1 }{ 2 }(a-b)$

5. jim_thompson5910

you should get -ax - bx in the numerator when you expand things out

6. hpfan101

Though I don't know how to get to that

7. hpfan101

Oh ok, I'll change that

8. jim_thompson5910

do you see how I got -ax ?

9. hpfan101

I'm assuming when you multiplied everything out by the conjugate. Though in my work, I don't see that. I must have made an error

10. hpfan101

Oh never mind I see what I did wrong

11. hpfan101

Can the equation be simplified further?

12. DanJS

Calc 1 is 5% Calc shortcuts and 95% ALgebraaaa

13. jim_thompson5910

I would factor out -x. Then multiply top and bottom by 1/x $\Large \frac{-ax-bx}{\sqrt{x^2-ax}+\sqrt{x^2+bx}}$ $\Large \frac{-x(a+b)}{\sqrt{x^2-ax}+\sqrt{x^2+bx}}$ $\Large \color{red}{\frac{1/x}{1/x}\times}\frac{-x(a+b)}{\sqrt{x^2-ax}+\sqrt{x^2+bx}}$ $\Large \frac{(1/x)*-x(a+b)}{(1/x)*\left(\sqrt{x^2-ax}+\sqrt{x^2+bx}\right)}$ $\Large \frac{-1(a+b)}{(1/x)*\sqrt{x^2-ax}+(1/x)*\sqrt{x^2+bx}}$ Do you see how to finish up?

14. hpfan101

Yes I do. Thank you!

15. jim_thompson5910

ok great

16. DanJS

17. DanJS

you doing the Delta-epsilon definition thing yet? I just reviewed that

18. hpfan101

Not yet. I just learned in class about the Intermediate Value Theorem. But thanks for the cheat sheet!

19. IrishBoy123

are you sure that is the right answer? $$\lim_{x \rightarrow \infty}(\sqrt{x^2-ax}-\sqrt{x^2+bx})$$ $$=\lim_{x \rightarrow \infty}x(\sqrt{1-\frac{a}{x}}-\sqrt{1+\frac{b}{x}})$$ using binomial expansion $$\sqrt{1+x} = 1 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16}x^3 - \frac{5}{128}x^4 + \frac{7}{256}x^5 - \cdots$$\ $$=lim_{x \rightarrow \infty}x( \{ 1 +\frac{1}{2}(-\frac{a}{x}) - \frac{1}{8}(-\frac{a}{x})^2 + \cdots \} - \{1 +\frac{1}{2}(\frac{b}{x}) - \frac{1}{8}(\frac{b}{x})^2\})$$ $$= -\frac{1}{2}(a+b)$$

20. IrishBoy123

i mean this |dw:1443186003890:dw|

21. DanJS

It has the negative in front, -(a+b)/2 From here $\Large \frac{-x(a+b)}{\sqrt{x^2-ax}+\sqrt{x^2+bx}}$ take an x out of both roots in bottom and cancel it with the top to get $\Large \frac{-(a+b)}{\sqrt{1-\frac{ a }{ x }}+\sqrt{1+\frac{ b }{ x }} }$ As x goes to infinity, those fractions go to zero a/x and b/x Left with the limit at infinity as $\Large \frac{-(a+b)}{\sqrt{1}+\sqrt{1}} = \frac{-(a+b)}{2}$

22. DanJS

** $\Large \lim_{x \rightarrow \infty } \frac{ 1 }{ x } = 0$

23. IrishBoy123

yes @DanJS , just a typo i expect:-)

24. IrishBoy123

thx

25. DanJS

what did you do, that looks like the taylor series expansion of that roots