0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ That's not the answer to the question I posted. $\endgroup$ – 5xum Mar 7 '16 at 12:34
  • 2
    $\begingroup$ 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
4
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Good catch! OP should edit the question... $\endgroup$ – 5xum Mar 7 '16 at 12:39
0
$\begingroup$

Hint:

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.