1. ## limit problem

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

2. Originally Posted by thangbe
$
=\lim _{x->0} \frac{\frac{1}{\sqrt{a+x+2}} - \frac{1}{\sqrt{a+x}}}{x}$

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 ....

3. Originally Posted by thangbe
$
=\lim _{x->0} \frac{\frac{1}{\sqrt{a+x+2}} - \frac{1}{\sqrt{a+x}}}{x}
$

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.

4. 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 ....

5. 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.

6. Originally Posted by thangbe
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}}

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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.

7. 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.

8. 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.

9. 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.

10. 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.

11. 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.

12. 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 $x \rightarrow 0$.

13. thank you so much for helping