Question

\[\lim_{x \rightarrow \infty}(\sqrt{x^2-ax}-\sqrt{x^2+bx}\]

Answer 1

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

Answer 2

wow am i wrong!

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

Though I don't know how to get to that

Answer 7

Oh ok, I'll change that

Answer 8

do you see how I got `-ax` ?

Answer 9

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

Answer 10

Oh never mind I see what I did wrong

Answer 11

Can the equation be simplified further?

Answer 12

Calc 1 is 5% Calc shortcuts and 95% ALgebraaaa

Answer 13

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?

Answer 14

Yes I do. Thank you!

Answer 15

ok great

Answer 16

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

Answer 17

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

Answer 18

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
\(\sqrt{1+x} = 1 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16}x^3 - \frac{5}{128}x^4 + \frac{7}{256}x^5 - \cdots\)\

\(=lim_{x \rightarrow \infty}x( \{ 1 +\frac{1}{2}(-\frac{a}{x}) - \frac{1}{8}(-\frac{a}{x})^2 + \cdots \} - \{1 +\frac{1}{2}(\frac{b}{x}) - \frac{1}{8}(\frac{b}{x})^2\})\)

\(= -\frac{1}{2}(a+b)\)

Answer 19

i mean this

Answer 20

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
\[\Large \frac{-(a+b)}{\sqrt{1}+\sqrt{1}} = \frac{-(a+b)}{2}\]

Answer 21

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

Answer 22

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

Answer 23

thx

Answer 24

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