Evaluate the limit:

$$\lim_{x\to4^-} \frac{\sqrt{3-\sqrt{5+x}}}{1-\sqrt{5-x}}$$

I multiplied numerator and denominator with their respective conjugates and got: $$\lim_{x\to4^-} \frac{1}{\sqrt{3+\sqrt{5+x}}}.\frac{\sqrt{4-x}}{x-4}(1+\sqrt{5-x})$$

Now it looks like this limit is undefined. Am I correct?

  • $\begingroup$ Don't post images of equation or text. That's what mathjax is for. $\endgroup$ – 5xum Mar 7 '16 at 12:23
  • $\begingroup$ Also, what do you get after multiplying with conjugates? $\endgroup$ – 5xum Mar 7 '16 at 12:24
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    $\begingroup$ That's not the answer to the question I posted. $\endgroup$ – 5xum Mar 7 '16 at 12:34
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    $\begingroup$ If you take the limit $x\rightarrow 4^{-}$ you get the indefinite form $0/0, \,$ the first derivative of the denomiator goes to $1/2$ but the first derivative of the numerator goes to $-\infty$. $\endgroup$ – gammatester Mar 7 '16 at 12:46

The limit doesn't exists since if $\;x\to 4^+\;$ then

$$\;5+x>9\implies\sqrt{5+x}>\sqrt9=3\implies 3-\sqrt{5+x}<0\implies\sqrt{3-\sqrt{5+x}}$$

isn't defined on $\;\Bbb R\;$ . Perhaps it should be a one-sided limit?

  • $\begingroup$ Good catch! OP should edit the question... $\endgroup$ – 5xum Mar 7 '16 at 12:39


$${\sqrt{4-x}\over x-4}={-1\over\sqrt{4-x}}$$

It should now be clear what happens as $x\to4^-$.


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