$\displaystyle
=\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 $\displaystyle f(a) = \frac{1}{\sqrt{a + 2}}$. Then, with a (perhaps disquieting) change from the usual notation ($\displaystyle x$ instead of $\displaystyle h$, $\displaystyle a$ instead of $\displaystyle x$) you can consider the derivative of $\displaystyle f(a)$ from first principles:
$\displaystyle f'(a) = \lim_{x \rightarrow 0} \frac{f(a + x) - f(a)}{x}$.
So the answer will be the derivative of f(a) with respect to a ....
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.
Note that $\displaystyle \frac{1}{\sqrt{a + x + 2}} - \frac{1}{\sqrt{a+2}} = \frac{\sqrt{a+2} - \sqrt{a+x+2}}{\sqrt{a+x+2} \, \sqrt{a+2}}$.
Therefore:
$\displaystyle \frac{\frac{1}{\sqrt{a + x + 2}} - \frac{1}{\sqrt{a+2}}}{x} = \frac{\sqrt{a+2} - \sqrt{a+x+2}}{x \sqrt{a+x+2} \, \sqrt{a+2}}$
$\displaystyle = \frac{(\sqrt{a+2} - \sqrt{a+x+2}) \cdot (\sqrt{a+2} + \sqrt{a+x+2})}{x \sqrt{a+x+2} \, \sqrt{a+2} \cdot (\sqrt{a+2} + \sqrt{a+x+2})}$
$\displaystyle = \frac{(a + 2) - (a + x + 2)}{x \sqrt{a+x+2} \, \sqrt{a+2} \cdot (\sqrt{a+2} + \sqrt{a+x+2})}$
$\displaystyle = \frac{-x}{x \sqrt{a+x+2} \, \sqrt{a+2} \cdot (\sqrt{a+2} + \sqrt{a+x+2})}$
$\displaystyle = \frac{-1}{\sqrt{a+x+2} \, \sqrt{a+2} \cdot (\sqrt{a+2} + \sqrt{a+x+2})}$.
Now take the limit $\displaystyle x \rightarrow 0$.