3
$\begingroup$

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

(I have nothing against using L'Hôpital's rule, but my book expects me to solve this without using it which is something I don't quite dominate yet, and so I will try for practice)

I tried the following:

$$\lim_{x \rightarrow 1}\frac{\sqrt x -1}{2-\sqrt{x+3}} \cdot \frac{2+\sqrt{x+3}}{2+\sqrt{x+3}} = \lim_{x \rightarrow 1}\frac{(\sqrt x -1)(2+\sqrt{x+3})}{4-(x+3)} = \lim_{x \rightarrow 1}\frac{2\sqrt x +\sqrt x \sqrt{x+3}-2-\sqrt{x+3}}{1-x} = ???$$

What do I do next?

  • 2
    Shouldn't have expanded the numerator What is $\frac{\sqrt{x}-1}{1-x}$? – 2017-02-18
  • 0
    @ThomasAndrews $-\frac{1}{\sqrt{x-1}}$? – 2017-02-18

2 Answers 2

5

In your last step, rather than expand out the numerator you should leave it alone, and instead note that $$ \frac{(\sqrt{x}-1)(2+\sqrt{x+3})}{1-x}=\frac{(\sqrt{x}-1)(2+\sqrt{x+3})}{(1-\sqrt{x})(1+\sqrt{x})}=-\frac{(2+\sqrt{x+3})}{1+\sqrt{x}} $$

  • 0
    I don't understand one thing though, why the minus sign on the last step? – 2017-02-18
  • 2
    $\dfrac{ \sqrt{x}-1 }{ 1-\sqrt{x}} = \dfrac{ \sqrt{x} - 1}{ - (-1 + \sqrt{x}} = \dfrac{ \sqrt{x} - 1}{ - (\sqrt{x} - 1)}$ – 2017-02-18
1

In that case when it is $\frac00$ you can do a double conjugation, thus multiplying both numerator and denominator by their respective conjugated quantity.

$\frac{\sqrt x-1}{2-\sqrt{x+3}}\times\frac{2+\sqrt{x+3}}{2+\sqrt{x+3}}\times\frac{\sqrt x+1}{\sqrt x+1}=\frac{(x-1)}{(1-x)}\times\frac{\sqrt x+1}{2+\sqrt{x+3}}=-1\times\frac{\sqrt x+1}{2+\sqrt{x+3}}\to-\frac12$

And you notice that $\frac{num}{den}=\frac{rationalized(num)}{rationalized(den)}\times\frac{conj(num)}{conj(den)}$

with $\frac{rationalized(num)}{rationalized(den)}$ carrying the undetermined $\frac00$ form but fortunately simplifying

and $\frac{conj(num)}{conj(den)}$ is not undetermined anymore.

All exercises of this type share this trick, so think about it :-).

  • 0
    Very interesting. This might be very useful to me. Thanks – 2017-02-18