• Oct 26th 2006, 01:37 AM
m777
• Oct 26th 2006, 02:34 AM
CaptainBlack
Quote:

Originally Posted by m777

Q1. find:

$\displaystyle \lim_{x \to \infty} [\sqrt{x^2+a x} -x]$

rewrite as:

$\displaystyle \lim_{x \to \infty} [x \sqrt{1+a/x} -x]$

or:

$\displaystyle \lim_{x \to \infty} x[ \sqrt{1+a/x} -1]$

Which may be rewritten:

$\displaystyle \lim_{x \to \infty} \frac{ \sqrt{1+a/x} -1}{1/x}$

and now L'Hopitals rule may be applied to give:

$\displaystyle \lim_{x \to \infty} \frac{ \sqrt{1+a/x} -1}{1/x}= \lim_{x \to \infty} \frac{ (1/2)(1+a/x)^{-1/2}(-a/x^2)}{-1/x^2}=$$\displaystyle \lim_{x \to \infty} (1+a/x)^{-1/2}(a/2)=a/2$

RonL
• Oct 26th 2006, 05:15 AM
nictan
Well for Qn 2 its quite stright forward

Just change all the x into (1)

then you will get

= 2.(1)²

= 2. (1) <--- (1)² = 1 hence it is = 2
• Oct 26th 2006, 06:36 AM
ThePerfectHacker
[QUOTE=ThePerfectHacker;25185]Question 1, Use the standard rationalization trick,

The function,
$\displaystyle f(x)=\sqrt{x^2+ax}-x$
And,
$\displaystyle g(x)=\frac{\sqrt{x^2+ax}-x}{1}\cdot \frac{\sqrt{x^2+ax}+x}{\sqrt{x^2+ax}+x}$
Agree for sufficiently large $\displaystyle x$, thus the limits are the same.

Working with the second function we find,
$\displaystyle \frac{x^2+ax-x^2}{\sqrt{x^2+ax}+x}$
Thus,
$\displaystyle \frac{ax}{\sqrt{x^2+ax}-x}$
For suffiently large, i.e., $\displaystyle x>0$ we factor,
$\displaystyle \frac{ax}{x\sqrt{1+a/x}+x}$
Factor again,
$\displaystyle \frac{ax}{x(\sqrt{1+a/x}+1)}$
Thus,
$\displaystyle \frac{a}{\sqrt{1+a/x}+1}$
Now,
$\displaystyle \lim_{x\to \infty}=\sqrt{1+a/x}=1$ limit compositions rule,
Thus, by limit composition rule again,
$\displaystyle \frac{a}{\sqrt{1+a/x}+1}\to \frac{a}{1+1}=a/2$
• Oct 26th 2006, 09:08 AM
Soroban
Hello, m777!

A slightly different approach to #1 . . .

Quote:

$\displaystyle 1)\;\lim_{x\to\infty}\left(\sqrt{x^2+ax} - x\right)$

Multiply top and bottom by $\displaystyle (\sqrt{x^2+ax} + x)$

. . $\displaystyle \frac{\sqrt{x^2+ax} - x}{1}\cdot\frac{\sqrt{x^2+ax} + x}{\sqrt{x^2+ax} + x} \;=\;\frac{(x^2 + ax) - x^2}{\sqrt{x^2+ax} + x} \;=\;\frac{ax}{\sqrt{x^2+ax} + x}$

Divide top and bottom by $\displaystyle x:\;\;\frac{\frac{ax}{x}}{\frac{\sqrt{x^2+ax}}{x} + \frac{x}{x}} \;=\;\frac{a}{\sqrt{\frac{x^2}{x^2}+\frac{a^2}{x^2 }} + 1} \;=\;\frac{a}{\sqrt{1 + \frac{a^2}{x^2}} + 1}$

Take the limit: .$\displaystyle \lim_{x\to\infty}\frac{a}{\sqrt{1 + \frac{a^2}{x^2}} + 1} \;=\;\frac{a}{\sqrt{1 + 0} + 1} \:=\:\frac{a}{2}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Quote:

$\displaystyle 3)\;f(x)\:=\:\left\{\begin{array}{ccc}\frac{x^2-a^2}{x-a} & 0 \leq x < a \\ a & x = a \\ 2a & x > a\end{array} \right\}$

Did you make a sketch?
Code:

            |             |    (a,2a)             |      o * * * * * *             |    *             |  *             | *    *(a,a)             *             |     --------+-------+--------------- -             |      a

Rather obvious that it's discontinuous at $\displaystyle x = a$, isn't it?

• Oct 26th 2006, 09:20 AM
CaptainBlack
Quote:

Originally Posted by Soroban
Hello, m777!

A slightly different approach to #1 . . .

Multiply top and bottom by $\displaystyle (\sqrt{x^2+ax} + x)$

. . $\displaystyle \frac{\sqrt{x^2+ax} - x}{1}\cdot\frac{\sqrt{x^2+ax} + x}{\sqrt{x^2+ax} + x} \;=\;\frac{(x^2 + ax) - x^2}{\sqrt{x^2+ax} + x} \;=\;\frac{ax}{\sqrt{x^2+ax} + x}$

Divide top and bottom by $\displaystyle x:\;\;\frac{\frac{ax}{x}}{\frac{\sqrt{x^2+ax}}{x} + \frac{x}{x}} \;=\;\frac{a}{\sqrt{\frac{x^2}{x^2}+\frac{a^2}{x^2 }} + 1} \;=\;\frac{a}{\sqrt{1 + \frac{a^2}{x^2}} + 1}$

Take the limit: .$\displaystyle \lim_{x\to\infty}\frac{a}{\sqrt{1 + \frac{a^2}{x^2}} + 1} \;=\;\frac{a}{\sqrt{1 + 0} + 1} \:=\:\frac{a}{2}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Did you make a sketch?
Code:

            |             |    (a,2a)             |      o * * * * * *             |    *             |  *             | *    *(a,a)             *             |     --------+-------+--------------- -             |      a
Rather obvious that it's discontinuous at $\displaystyle x = a$, isn't it?

Unless a=0.

RonL
• Oct 26th 2006, 09:32 AM
topsquark
Is a function that is only defined for a single point continuous?

-Dan
• Oct 26th 2006, 09:42 AM
ThePerfectHacker
Quote:

Originally Posted by topsquark
Is a function that is only defined for a single point continuous?

No! by definition,
$\displaystyle \lim_{x\to c}f(x)=f(c)$
Now in order to take a limit we need to find a $\displaystyle \delta>0$ such that,
$\displaystyle |x-c|<\delta$ is in the domain, that means,
$\displaystyle c-\delta<x<c+\delta$
Is some open interval.
That means the function needs to be definied on some open interval about $\displaystyle x=c$ which it fails heir.
• Oct 26th 2006, 10:29 AM
m777