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How to find $$\lim_{x \to 1} \frac{\sqrt{x}+\sqrt{x+3}-3}{x+x^2-2}$$

My Try :

$$x^2+x-2=(x-1)(x+2)$$

$$\lim_{x \to 1} \frac{\sqrt{x}+\sqrt{x+3}-3}{(x-1)(x+2)}\cdot \frac{(\sqrt{x}+\sqrt{x+3})+3}{(\sqrt{x}+\sqrt{x+3})+3}=\\ =\lim_{x \to 1} \frac{2x+2\sqrt{x(x+3)}-6}{(x-1)(x+2)((\sqrt{x}+\sqrt{x+3})+3)}=\\\lim_{x \to 1} \frac{2(x-3+\sqrt{x(x+3)})}{(x-1)(x+2)((\sqrt{x}+\sqrt{x+3})+3)} $$

Now what ?

Almot1960
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  • Since that $x+2$ has got nothing to do with the actual limit problem, pull it upfront and leave that for the end (by limit laws). At least you have one factor less to worry about in the algebraic forest. – imranfat Oct 16 '17 at 18:28

3 Answers3

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First factor $x^2+x^2-2=(x-1)(x+2)$. $$\lim_{x \to 1} \frac{\sqrt{x}+\sqrt{x+3}-3}{(x-1)(x+2)}=$$

$$=\lim_{x \to 1} \frac{\sqrt{x}+\sqrt{x+3}-3}{x-1}\lim_{x\to 1}\frac{1}{x+2}=$$

$$=\left(\lim_{x\to 1}\left(\frac{\sqrt{x}-1}{x-1}+\frac{\sqrt{x+3}-2}{x-1}\right)\right)\cdot \frac{1}{3}$$

Hint: multiply the numerator and denominator of $\frac{\sqrt{x}-1}{x-1}$ by $\sqrt{x}+1$ and of $\frac{\sqrt{x+3}-2}{x-1}$ by $\sqrt{x+3}+2$.

user236182
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HINT: write $$\frac{2(x-3+\sqrt{x(x+3)})(x-3-\sqrt{x(x+3)})}{(x-1)(x+2)(\sqrt{x}+\sqrt{x+3}+3)(x-3-\sqrt{x(x+3)}}$$ and this is equal $$\frac{-9(x-1)}{(x-1)(x+2)(\sqrt{x}+\sqrt{x+3}+3)(x-3-\sqrt{x(x+3)})}$$

Dr. Sonnhard Graubner
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Say $t=\sqrt{x}$, then

\begin{eqnarray} % \nonumber to remove numbering (before each equation) \lim_{t \to 1} \frac{t+\sqrt{t^2+3}-3}{t^4+t^2-2} &= &\lim_{t \to 1} \frac{t-1+\sqrt{t^2+3}-2}{(t^2+2)(t^2-1)} \\ &=& \lim_{t \to 1} \frac{1}{(t^2+2)(t+1)} + \frac{\sqrt{t^2+3}-2}{(t^2+2)(t^2-1)}\\ &=& {1\over 6}+\lim_{t \to 1} \frac{t^2-1}{(t^2+2)(t^2-1)(\sqrt{t^2+3}+2)}\\ &=& {1\over 6}+ {1\over 12}\\ &=& {1\over 4} \end{eqnarray}

nonuser
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