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$\lim_{x\to2}\frac{\sqrt{1+\sqrt{x+2}-\sqrt3}}{x-2}$

My try:
$\lim_{x\to2}\frac{\sqrt{1+\sqrt{x+2}-\sqrt3}}{x-2}=$
$\lim_{x\to2}\frac{\sqrt{1+\sqrt{x+2}-\sqrt3}}{x-2}\times\frac{\sqrt{1+\sqrt{x+2}+\sqrt3}}{\sqrt{1+\sqrt{x+2}+\sqrt3}}=$
$\lim_{x\to2}\frac{\sqrt{x+2\sqrt{x+2}}}{(x-2)\sqrt{1+\sqrt{x+2}+\sqrt3}}$

I am stuck here.

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    Are you sure you copied the question correctly? It would be a more interesting limit problem if the numerator was $\sqrt{1+\sqrt{x+2}}-\sqrt3$ instead of $\sqrt{1+\sqrt{x+2}-\sqrt3}$ (mind the outer square root). – 2017-02-13
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    I elaborated on the alternative limit in another answer. – 2017-02-13

4 Answers 4

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Note that $$\lim_{x\to2}\frac{\sqrt{1+\sqrt{x+2}-\sqrt3}}{x-2}$$ The denominator goes to zero, but the numerator goes to $\sqrt{3-\sqrt{3}}$.

Thus, we know that the limit that you are trying to evaluate does not exist, and the limit goes to $-\infty$ and $\infty$ respectively as $x \to 2$.

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    Just to add that one-directional limit does exist (from left or from right) – 2017-02-13
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    @rbm Do you think my edit covers it? – 2017-02-13
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    Yep, looks good. – 2017-02-13
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$$\lim\limits_{x\to 2^{\pm}}\frac{\sqrt{1+\sqrt{x+2}-\sqrt{3}}}{x-2} = \left[\frac{\sqrt{3-\sqrt{3}}}{0^{\pm}}\right]=\pm\infty$$

As you can see, the left- and right-side limits are different, so the limit $\lim\limits_{x\to 2}\frac{\sqrt{1+\sqrt{x+2}-\sqrt{3}}}{x-2}$ does not exists.

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    True, but not what OP\s asked. – 2017-02-13
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    @rbm I've edited my post. And, actually, OP didn't asked about anything at all ;) – 2017-02-13
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    That's true, i assumed it was implied. +1! – 2017-02-13
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Following my comment, the question would make "more sense" (seeing that you know the method of multiplying with a conjugate expression) if the numerator was $\sqrt{1+\sqrt{x+2}}-\sqrt3$.

If you did copy the question correctly, then it might have been a bit of a trick question since you don't need that method; it is not an indeterminate form. Other answers (by S.C.B. and Jaroslaw Matlak) explain the way to go.

If the numerator is as I suggested, then your idea of multiplying with the conjugate expression is fine and you can apply that trick twice in order to cancel out a common factor $\color{red}{x-2}$: $$\begin{align}\lim_{x\to2}\frac{\sqrt{1+\sqrt{x+2}}-\sqrt3}{x-2} & = \lim_{x\to2}\frac{\sqrt{1+\sqrt{x+2}}-\sqrt3}{x-2}\color{blue}{\frac{\sqrt{1+\sqrt{x+2}}+\sqrt3}{\sqrt{1+\sqrt{x+2}}+\sqrt3}} \\[7pt] & = \lim_{x\to2}\frac{1+\sqrt{x+2}-3}{\left(x-2\right)\left(\sqrt{1+\sqrt{x+2}}+\sqrt3\right)} \\[7pt] & = \lim_{x\to2}\frac{\sqrt{x+2}-2}{\left(x-2\right)\left(\sqrt{1+\sqrt{x+2}}+\sqrt3\right)}\color{blue}{\frac{\sqrt{x+2}+2}{\sqrt{x+2}+2}} \\[7pt] & = \lim_{x\to2}\frac{\color{red}{x-2}}{\color{red}{\left(x-2\right)}\left(\sqrt{1+\sqrt{x+2}}+\sqrt3\right)\left(\sqrt{x+2}+2\right)} \\[7pt] & = \lim_{x\to2}\frac{1}{\left(\sqrt{1+\sqrt{x+2}}+\sqrt3\right)\left(\sqrt{x+2}+2\right)}= \frac{1}{8\sqrt{3}} \\[7pt] \end{align}$$ Perhaps this still helps.

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Following StackTD's comment

let $\sqrt{1+\sqrt{x+2}}-\sqrt3=y\implies y\to0$

and $1+\sqrt{x+2}=(y+\sqrt3)^2=y(y+2\sqrt3)+3$

$$x+2=(y^2+2\sqrt3y+2)^2=y^2(y+2\sqrt3)^2+4y(y+2\sqrt3)+4$$

$$\lim_{x\to2}\frac{\sqrt{1+\sqrt{x+2}}-\sqrt3}{x-2}=\lim_{y\to0}\dfrac y{y^2(y+2\sqrt3)^2+4y(y+2\sqrt3)}=?$$