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$$\lim_{x \to 2}⁡{\sqrt{x+2}- \sqrt{x+7}+1\over x-2}$$

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    It's 0/0 so apply L'Hopital – 2017-01-04
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    What an interesting expression. I will gaze at it – 2017-01-04
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    You should get in the habit of stating your question as a... question. Do you want some hint to evaluate this? – 2017-01-04

3 Answers 3

2

L'Hospital's rule is super easy:

$$\lim_{x\to2}\frac{\sqrt{x+2}- \sqrt{x+7}+1}{x-2}=\lim_{x\to2}\frac1{2\sqrt{x+2}}-\frac1{2\sqrt{x+7}}=\frac1{12}$$

2

Without l'Hospital and twice multiplying by the conjugate of the appropiate expression:

$$\lim_{x \to 2}⁡{\sqrt{x+2}+1- \sqrt{x+7}\over x-2}=\frac{x+2+1-x-7+2\sqrt{x+2}}{(x-2)(\color{red}{\sqrt{x+2}+1+\sqrt{x+7}})}=$$

$$=-2\frac{\left(2-\sqrt{x+2}\right)}{(x-2)(\sqrt{x+2}+1+\sqrt{x+7})}=-2\frac{4-x-2}{(x-2)(\sqrt{x+2}+1+\sqrt{x+7})\color{red}{(2+\sqrt{x+2})}}=$$

$$=\frac2{(\sqrt{x+2}+1+\sqrt{x+7})(2+\sqrt{x+2})}\xrightarrow[x\to2]{}\frac2{(2+1+3)(2+2)}=\frac1{12}$$

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    Finely done. +1 I just dislike the amount of errors I usually make doing it this way T_T – 2017-01-04
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    @SimpleArt You and I both, believe me. For what is worth, if I had to do this thing I wouldn't even try to my way: I'd go l'Hospital without thinking twice. – 2017-01-04
0

We use L'Hopital's rule:

$$\lim_{x \to a} \frac{f(x)}{g(x)}=\lim_{x \to a} \frac{f'(x)}{g'(x)} \tag{1}$$

Let $f(x)=\sqrt{x+2}-\sqrt{x+7}+1$ and $g(x)=x+2$.

We evaluate the derivatives of these functions:

$f'(x)=\frac{1}{2\sqrt{x+2}}-\frac{1}{2\sqrt{x+7}}$ and $g'(x)=1$

Substitute into equation $(1)$ (Apply L'Hopital's rule).

$$\lim_{x \to 2} \frac{\sqrt{x+2}-\sqrt{x+7}+1}{x+2}=\lim_{x \to 2} \frac{\frac{1}{2\sqrt{x+2}}-\frac{1}{2\sqrt{x+7}}}{1}$$

Substitute $x=2$.

$$\frac{1}{4}-\frac{1}{6}=\frac{1}{12}$$

Hence,

$$\lim_{x \to 2} \frac{\sqrt{x+2}-\sqrt{x+7}+1}{x+2}=\frac{1}{12}$$

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    This is, as far as I can see, exactly the very same Simple Art posted some 6 minutes ago... – 2017-01-04