limit problem

**thangbe** · February 28th 2009, 09:54 PM

$ =\lim _{x->0} \frac{\frac{1}{\sqrt{a+x+2}} - \frac{1}{\sqrt{a+x}}}{x} $

could you please help me to solve this problem please. Thank you.

**mr fantastic** · March 1st 2009, 03:36 AM

Originally Posted by thangbe

$ =\lim _{x->0} \frac{\frac{1}{\sqrt{a+x+2}} - \frac{1}{\sqrt{a+x}}}{x}$

could you please help me to solve this problem please. Thank you.

Several approaches are possible.

There's a simple one if you know differential calculus ..... you can consider the function $f(a) = \frac{1}{\sqrt{a + 2}}$ . Then, with a (perhaps disquieting) change from the usual notation ( $x$ instead of $h$ , $a$ instead of $x$ ) you can consider the derivative of $f(a)$ from first principles:

$f'(a) = \lim_{x \rightarrow 0} \frac{f(a + x) - f(a)}{x}$ .

So the answer will be the derivative of f(a) with respect to a ....

**HallsofIvy** · March 1st 2009, 04:26 AM

Originally Posted by thangbe

$ =\lim _{x->0} \frac{\frac{1}{\sqrt{a+x+2}} - \frac{1}{\sqrt{a+x}}}{x} $

could you please help me to solve this problem please. Thank you.

If you set x= 0, the numerator is $\frac{1}{\sqrt{a+ 2}}- \frac{1}{\sqrt{a}}$ which is clearly NOT equal to 0 but the denominator is 0. Obviously this limit does not exist.

**mr fantastic** · March 1st 2009, 04:30 AM

Originally Posted by HallsofIvy

If you set x= 0, the numerator is $\frac{1}{\sqrt{a+ 2}}- \frac{1}{\sqrt{a}}$ which is clearly NOT equal to 0 but the denominator is 0. Obviously this limit does not exist.

Dang it. I misread the red x for a 2:

] $ =\lim _{x->0} \frac{\frac{1}{\sqrt{a+x+2}} - \frac{1}{\sqrt{a+{\color{red}x}}}}{x} $

Hmmmmmm ... I bet that red x is meant to be a 2 ....

**thangbe** · March 1st 2009, 05:06 PM

the limit is exists and the answer is.
$ =\lim _{x->0} \frac{\frac{1}{\sqrt{a+x+2}} - \frac{1}{\sqrt{a+x}}}{x} =\frac {-1}{2{(a+2)}^\frac{3}{2}} $

but I don't know how to show for it.

**mr fantastic** · March 1st 2009, 05:31 PM

Originally Posted by thangbe

the limit is exists and the answer is.
$ 0} \frac{\frac{1}{\sqrt{a+x+2}} - \frac{1}{\sqrt{a+x}}}{x} =\frac {-1}{2{(a+2)}^\frac{3}{2}}

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=\lim _{x->0} \frac{\frac{1}{\sqrt{a+x+2}} - \frac{1}{\sqrt{a+x}}}{x} =\frac {-1}{2{(a+2)}^\frac{3}{2}}

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but I don't know how to show for it.

So you have mistyped the question as I suspected - see post #4.

One way of getting the answer is given in post #2. Study it. If you do not understand that method, then you will need to say so. There are several other methods that can be used.

**thangbe** · March 1st 2009, 06:33 PM

I understand in your post #2 that why I get that answer. However my teacher didn't allows us to use that menthod

thank you for you help sir.

**matheagle** · March 1st 2009, 06:41 PM

Clearly there is a typo.
But once that is resolved, my favourite way to solve these type of limits is by multiplying by the conjugate. Not many people teach that these days.

**mr fantastic** · March 1st 2009, 06:56 PM

Originally Posted by matheagle

Clearly there is a typo.
But once that is resolved, my favourite way to solve these type of limits is by multiplying by the conjugate. Not many people teach that these days.

I was going to do that but decided it was too much work to type it up.

**matheagle** · March 1st 2009, 07:05 PM

I agree. I'm helping someone right now.
But substituting the math and typing tex at the same is tough for me.
I'm afraid I'll do a stupid calculation and I cannot edit after 15 minutes.
Leaving forever with a dumb math error.
That would be embarassing.
I really like the conjugate technique.

**thangbe** · March 1st 2009, 07:18 PM

uhm... could you just simply tell me the procedure.? or just give a simple example. So that I can figure it out.... I was stuck in how to get x out of the squareroot value.

**mr fantastic** · March 2nd 2009, 01:49 AM

Originally Posted by thangbe

I understand in your post #2 that why I get that answer. However my teacher didn't allows us to use that menthod

thank you for you help sir.

Note that $\frac{1}{\sqrt{a + x + 2}} - \frac{1}{\sqrt{a+2}} = \frac{\sqrt{a+2} - \sqrt{a+x+2}}{\sqrt{a+x+2} \, \sqrt{a+2}}$ .

Therefore:

$\frac{\frac{1}{\sqrt{a + x + 2}} - \frac{1}{\sqrt{a+2}}}{x} = \frac{\sqrt{a+2} - \sqrt{a+x+2}}{x \sqrt{a+x+2} \, \sqrt{a+2}}$

$= \frac{(\sqrt{a+2} - \sqrt{a+x+2}) \cdot (\sqrt{a+2} + \sqrt{a+x+2})}{x \sqrt{a+x+2} \, \sqrt{a+2} \cdot (\sqrt{a+2} + \sqrt{a+x+2})}$

$= \frac{(a + 2) - (a + x + 2)}{x \sqrt{a+x+2} \, \sqrt{a+2} \cdot (\sqrt{a+2} + \sqrt{a+x+2})}$

$= \frac{-x}{x \sqrt{a+x+2} \, \sqrt{a+2} \cdot (\sqrt{a+2} + \sqrt{a+x+2})}$

$= \frac{-1}{\sqrt{a+x+2} \, \sqrt{a+2} \cdot (\sqrt{a+2} + \sqrt{a+x+2})}$ .

Now take the limit $x \rightarrow 0$ .

**thangbe** · March 2nd 2009, 06:07 AM

thank you so much for helping

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