How can I evaluate $\lim_{x\to7}\frac{\sqrt{x}-\sqrt{7}}{\sqrt{x+7}-\sqrt{14}}$?

Question

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?

Answer 1

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.

Answer 2

Hint: Multiply top and bottom by $\left(\sqrt{x}+7\right)\left(\sqrt{x+7}+14\right)$.

Answer 3

You can divide top and bottom by $x-7$ and identify your limit as the quotient of two derivatives at $x=7$.

Answer 4

Hint: Multiply by the conjugate.

Answer 5

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 $x\to 7$.