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I need to solve $$\lim_{x\to2^{+}}\frac{\sqrt{x+7}-3}{\sqrt{x^{2}+5}-x-1}$$ without using L'hopital or taylor. tried Conjugate multiplication to no end. Any ideas?

user114138
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  • a general advice: if you see a root in the numerator, minus or plus something else, start by trying to get an expression of the form $x^2 - y^2$ – Alex Nov 22 '15 at 10:39

2 Answers2

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You should give conjugate multiplication a second chance: $$\begin{align} \frac{\sqrt{x+7}-3}{\sqrt{x^{2}+5}-x-1}&=\frac{(\sqrt{x+7}-3)(\sqrt{x+7}+3)(\sqrt{x^{2}+5}+x+1)}{(\sqrt{x^{2}+5}-x-1)(\sqrt{x^{2}+5}+x+1)(\sqrt{x+7}+3)}\\ &=\frac{(x-2)(\sqrt{x^{2}+5}+x+1)}{(x^2+5-(x+1)^2)(\sqrt{x+7}+3)}\\ &=-\frac{\sqrt{x^{2}+5}+x+1}{2(\sqrt{x+7}+3)}. \end{align}$$

Julián Aguirre
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My preferred is to use $\sqrt{1+h}=1+{h\over 2}+o(h)$

In our case, order one is enough and we need first to write $x=2+h$. The fraction transforms as follows

$${3\left(\sqrt{1+{h\over 9}}-1\right)\over 3\left(\sqrt{1+{4h\over 9}+{h^2\over 9}}-{h\over 3}-1\right)}$$

and this simplifies in one step into

$${{h\over 18}+o(h)\over -{h\over 9}+o(h)}=-{1\over 2}+o(1)$$

And the limit is $-{1\over 2}$

marwalix
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