How to evaluate $\lim _{x\to 2^+}\left(\frac{\sqrt{x^2-4}+\sqrt{x}-\sqrt{2}}{\sqrt{x-2}}\right)$ (without L'Hopital)?

Question

I am trying to evaluate the following limit:

$$ \lim _{x\to 2^+}\left(\frac{\sqrt{x^2-4}+\sqrt{x}-\sqrt{2}}{\sqrt{x-2}}\right)$$

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Approach #1

$ \frac{\sqrt{x^2-4}+\sqrt{x}-\sqrt{2}}{\sqrt{x-2}} = \\ \sqrt{x+2} + \frac{\sqrt{x}-\sqrt{2}}{\sqrt{x-2}} \cdot \frac{\sqrt{x}+\sqrt{2}}{\sqrt{x}+\sqrt{2}} =\\ \sqrt{x+2} + \frac{x-2}{\sqrt{x^2-2x}+\sqrt{2x-4}} =\\ \sqrt{x+2} + \frac{x-2}{\sqrt{x^2-2x}+\sqrt{2x-4}}\cdot\frac{\sqrt{x^2-2x}-\sqrt{2x-4}}{\sqrt{x^2-2x}-\sqrt{2x-4}} = \\ \sqrt{x+2} + \frac{\sqrt{x^2-2x}-\sqrt{2x-4}}{x-2} $

But I still end up at the indefinite form $\frac12 + \frac{0}{\infty}$

My Approach #2

$\frac{\sqrt{x^2-4}+\sqrt{x}-\sqrt{2}}{\sqrt{x-2}} = \frac{\sqrt{x^2-4}+\sqrt{x}-\sqrt{2}}{\sqrt{x-2}} \cdot\frac{\sqrt{x^2-4}-(\sqrt{x}-\sqrt{2})}{\sqrt{x^2-4}-(\sqrt{x}-\sqrt{2})}= \frac1{\sqrt{x-2}}\frac{x^2-x+2\sqrt{2x}-2}{\sqrt{x^2-4}-(\sqrt{x}-\sqrt{2})}$

Which also seems to be a dead end.

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Any ideas on how to evaluate this?

Answer 1

As regards your first approach, from the first line, as $x\to 2^+$, $$\frac{\sqrt{x^2-4}+\sqrt{x}-\sqrt{2}}{\sqrt{x-2}}=\underbrace{\color{blue}{\sqrt{x+2}}}_{\to 2}+\underbrace{\frac{\sqrt{x}-\sqrt{2}}{\sqrt{x-2}}}_{0/0}.$$ Now note that the $0/0$ indeterminate term is $$\frac{\sqrt{x}-\sqrt{2}}{\sqrt{x-2}}=\frac{\sqrt{x}-\sqrt{2}}{\sqrt{x-2}}\cdot \frac{\sqrt{x}+\sqrt{2}}{\sqrt{x}+\sqrt{2}}= \frac{x-2}{\color{blue}{\sqrt{x-2}(\sqrt{x}+\sqrt{2})}}=\frac{\sqrt{x-2}}{\sqrt{x}+\sqrt{2}}\to 0.$$ Therefore the required limit is $2$.

Answer 2

How about $$\frac{\sqrt x-\sqrt2}{\sqrt{x-2}} =\frac{x-2}{\sqrt{x-2}(\sqrt x+\sqrt2)} =\frac{\sqrt{x-2}}{\sqrt x+\sqrt2}\to0 $$ as $x\to2^+$. You've successfully tackled the rest.

Answer 3

For $x\ne 2$,$$\sqrt{x^2-4}=\sqrt{x+2}\sqrt{x-2}$$ and $$\sqrt{x-2}=\sqrt{\sqrt x+\sqrt2}\sqrt{\sqrt x-\sqrt2}$$

so that by simplification

$$\frac{\sqrt{x^2-4}+\sqrt{x}-\sqrt{2}}{\sqrt{x-2}} =\frac{\sqrt{x+2}\sqrt{\sqrt x+\sqrt2}+\sqrt{\sqrt x-\sqrt2}}{\sqrt{\sqrt x+\sqrt2}}.$$

The limit is $\sqrt{2+2}$.

Answer 4

For $y=x-2$, you have $$\frac{\sqrt{x^2-4}+\sqrt{x}-\sqrt{2}}{\sqrt{x-2}}=\frac{\sqrt{(y+2)^2-4}+\sqrt{y+2}-\sqrt{2}}{\sqrt{y}} = \frac{\sqrt{y^2+4y}+\sqrt{y+2}-\sqrt{2}}{\sqrt{y}}$$

When $x$ tends to $2$, $y$ tends to $0$, so you have $$\frac{\sqrt{y^2+4y}+\sqrt{y+2}-\sqrt{2}}{\sqrt{y}} = \frac{2\sqrt{y}\sqrt{1 + \frac{y}{4}}+\sqrt{2}\sqrt{1 + \frac{y}{2}}-\sqrt{2}}{\sqrt{y}}$$ $$ = \frac{2\sqrt{y}(1+ \frac{y}{8} + o \left( y\right))+\sqrt{2}(1 + \frac{y}{4} + o(y))-\sqrt{2}}{\sqrt{y}}$$ $$= 2 \left(1+ \frac{\sqrt{y}}{8} + o \left( \sqrt{y}\right)\right)+\sqrt{2}\left( \frac{\sqrt{y}}{4} + o(\sqrt{y})\right)$$ $$=2 + \sqrt{y} \left( \frac{1+\sqrt{2}}{4} \right) + o \left( \sqrt{y}\right)$$

Therefore the limit when $y$ tends to $0$ is equal to $2$.

Answer 5

$$\lim _{x\to 2}\left(\frac{\sqrt{x^2-4}+\sqrt{x}-\sqrt{2}}{\sqrt{x-2}}\right) = \lim _{x\to 2}\left(\sqrt{x+2}+\frac{\sqrt{x}-\sqrt{2}}{\sqrt{x-2}}\right) = \lim _{x\to 2}\left(\sqrt{x+2}+\frac{\sqrt{x-2}}{\sqrt{x}+\sqrt{2}}\right)$$

Answer 6

Since the limit is defined for $x\to 2^+$, let $y^2=x-2\to 0$ then

$$\lim _{x\to 2}\left(\frac{\sqrt{x^2-4}+\sqrt{x}-\sqrt{2}}{\sqrt{x-2}}\right)=\lim _{y\to 0}\left(\frac{\sqrt{y^2(y^2+4)}+\sqrt{y^2+2}-\sqrt{2}}{y}\right)=$$

$$\lim _{y\to 0}\left(\frac{\sqrt{y^2(y^2+4)}+\sqrt{y^2+2}-\sqrt{2}}{y}\right)=\lim _{y\to 0}\frac{\sqrt{y^2(y^2+4)}}{y}+\lim _{y\to 0}\frac{\sqrt{y^2+2}-\sqrt{2}}{y}=$$

$$=\lim _{y\to 0}\sqrt{(y^2+4)}+\lim _{y\to 0}\frac{\sqrt{y^2+2}-\sqrt{2}}{y}=2+0=2$$

indeed by definition of derivative

$$f(y)=\sqrt{y^2+2} \implies f'(y)=\frac{2y}{2\sqrt{y^2+2}}$$

$$\lim _{y\to 0}\frac{\sqrt{y^2+2}-\sqrt{2}}{y}=f'(0)=0$$