How do I find $\underset{x\rightarrow\infty }{\lim}A_{x}=x-\sqrt{x^{2}-4x}$

Question

$$\underset{x\rightarrow\infty }{\lim}A_{x}=x-\sqrt{x^{2}-4x}$$

I know that the answer is $2$, but i cant seem to manipulate the equation to get it out?

Hint: Multiply by the conjugate surd, i.e. $\frac{x+\sqrt{x^2-4x}}{+\sqrt{x^2-4x}} = 1$ — Inazuma, Mar 30 '16 at 11:49
Do you mean to write that $A_x= x-\sqrt{x^{2}-4x}$ and you wish to find the limit as $A_x$ approaches infinity? As the question stands right now, the notation is very ambiguous. — zz20s, Mar 30 '16 at 11:56
zz20s, i think so? as in, a sequence $A_{x}$ where $x\in \mathbb{N}$ and the $A_{x}$ term is defined by $x-\sqrt{x^{2}-4x}$ and we want to find the limit as $x\rightarrow \infty $. how was the notation ambiguous? not saying your wrong just that I'd like to fix it if possible... — Lincoln77, Mar 30 '16 at 12:01
Better: "Find $\lim_{x \to\infty} A(x)$, where $A(x)=\dots$." Or simply "Find $\lim_{x \to\infty} (\dots)$." — Hans Lundmark, Mar 30 '16 at 12:38
(By the way, did you try searching before posting? This must have been answered hundreds of times already...) — Hans Lundmark, Mar 30 '16 at 12:40
@Lincoln77, See also : http://math.stackexchange.com/questions/1721628/limit-evaluation-for-infinity-infinity-sqrtx1-sqrtx/1721657#1721657 — lab bhattacharjee, Mar 31 '16 at 11:41

score 6 · Accepted Answer · answered Mar 30 '16 at 11:55

6

$$\underset{x\rightarrow\infty }{\lim}x-\sqrt{x^{2}-4x}\cdot \color{blue}{\frac{x+\sqrt{x^2-4x}}{x+\sqrt{x^2-4x}}}$$

$$=\lim_{x\to \infty} \frac {4x}{x+\sqrt{x^2-4x}}$$

Dvided by $x$

$$=\lim_{x\to \infty} \frac {4}{1+\underbrace{\sqrt{1-4/x}}_{\to 1}}=\frac 4 2=\color{red}2$$

answered Mar 30 '16 at 11:55

3SAT

score 2 · Answer 2 · answered Mar 30 '16 at 11:54

2

Follow Inazuma's hint and then reduce by dividing by x to see what happens. Of course you need to make assumption that x is not zero.

answered Mar 30 '16 at 11:54

plktrautman

score 1 · Answer 3 · answered Mar 30 '16 at 12:05

1

Let $x>4,$ then $x-\sqrt {x^2-4x}= x-x\sqrt {1-4/x}= x-x( 1-2/x+o(1/x))=2+o(1)$

We used $\sqrt {1+u}= 1+1/2u+o(u)$.

answered Mar 30 '16 at 12:05

Thomas

score 1 · Answer 4 · answered Mar 30 '16 at 14:04

As $x^2-4x=(x-2)^2-2^2,$ let $x-2=\csc2t$

As $x\to\infty,2t\to0^+$

$$\lim_{x\to\infty}(x-\sqrt{x^2-4x})=2+\lim_{t\to0^+}(\csc2t-\cot2t)=2+\lim_{t\to0^+}\dfrac{1-\cos2t}{\sin2t}$$

$$\lim_{t\to0^+}\dfrac{1-\cos2t}{\sin2t}=\lim_{t\to0^+}\dfrac{2\sin^2t}{2\sin t\cos t}=\lim_{t\to0^+}\tan t=?$$ as $\sin t\ne0$ as $t\ne0$ as $t\to0$

4 Answers4