How can I evaluate $\lim_{x\to7}\frac{\sqrt{x}-\sqrt{7}}{\sqrt{x+7}-\sqrt{14}}$? I need to evaluate the following limit, if it exists:
$$\lim_{x\to7}\frac{\sqrt{x}-\sqrt{7}}{\sqrt{x+7}-\sqrt{14}}$$
How can I solve it without using differentiation or L'Hôpital?
 A: Hint: Multiply top and bottom by $\left(\sqrt{x}+7\right)\left(\sqrt{x+7}+14\right)$. 
A: We can get a nice function whose limit will not be indeterminate, by multiplying numerator and denominator each by the conjugate of  $\sqrt{x + 7} - \sqrt{14}$:
$$\lim_{x\to7}\frac{\sqrt{x}-\sqrt{7}}{\sqrt{x+7}-\sqrt{14}}\cdot\dfrac{(\sqrt{x+7} + \sqrt{14})}{(\sqrt{x+7} + \sqrt{14})}$$
In the denominator, we obtain a difference of squares:
$(\sqrt{x + 7} - \sqrt{14})(\sqrt{x + 7} + \sqrt{14})\; =\; (x + 7) -14 \; =\; x-7 $ 
$$\lim_{x\to7}\frac{(\sqrt x-\sqrt 7)(\sqrt{x+7} + \sqrt{14})}{(x - 7)}$$
Hmmm...this doesn't seem to help us any, since both the numerator AND denominator will still evaluate to $0$. 
But wait! We can view the term in the denominator as a difference of squares: $$(x - 7) = (\sqrt x)^2 - (\sqrt 7)^2 = (\sqrt{x} - \sqrt 7)(\sqrt x + \sqrt 7)$$
That gives us:
$$\lim_{x\to7}\dfrac{(\sqrt x-\sqrt 7)(\sqrt{x+7} + \sqrt {14})}{(\sqrt{x} - \sqrt 7)(\sqrt x + \sqrt 7)}$$
Canceling the common factor in numerator and denominator, we have (provided $x \neq 7$):
$$\lim_{x\to7}\dfrac{(\sqrt{x+7} + \sqrt {14})}{(\sqrt x + \sqrt 7)} = \frac {2\sqrt{14}}{2 \sqrt 7} = \sqrt{\frac {14}{7}} = \sqrt 2$$
which can now be evaluated without problems. We are taking the limit as $x \to 7$ after all, and so, while the original function is undefined at $x = 7$, the limit as $x \to 7$ indeed exists.
A: Note that  $$(\sqrt{x+7}-\sqrt{14})(\sqrt{x+7}+\sqrt{14})=(x+7)-14=x-7=(\sqrt x-\sqrt 7)(\sqrt x+\sqrt 7),$$hence
$$ \frac{\sqrt x-\sqrt 7}{\sqrt{x+7}-\sqrt{14}}=\frac{\sqrt{x+7}+\sqrt{14}}{\sqrt x+\sqrt 7}.$$
For the latter we obtain $\frac{\sqrt{14}+\sqrt{14}}{\sqrt7+\sqrt 7}=\sqrt 2$ as $x\to 7$.
A: You can divide top and bottom by $x-7$ and identify your limit as the quotient of two derivatives at $x=7$.
A: Hint: Multiply by the conjugate.
