Question

Finding the value of limit

\[\Large \lim_{x \to 3} \frac{\sqrt{x+4}-\sqrt{2x+1}}{\sqrt{3x+1}-\sqrt{x+7}}\]

Answer 1

I would use l'Hôpital's Rule in this case.

Answer 2

other than l'Hôpital's Rule?

Answer 3

I started a calculation, but got stuck :(
Therefore, I went to the emergency exit: l'Hôpital :)

Answer 4

I've tried to multiply by the conjugates of the numerator and got stuck, then trying to multiply by the conjugates of the denumerator and got stuck too...

am I allowed to multiply twice? by the conjugates of the numerator and denumerator?

Answer 5

As long as you multiply num. and den. with the same number, anything goes.
I also tried ot, but didn't get any further. That's why I came up with l'Hôpital.
That always feels a bit like cheating, though...

Answer 6

it seems doesn't work...

Answer 7

multiply both by conjugate ... cancel (x-3) from both of them.

Answer 8

That works!
\[\frac{ (\sqrt{x+4}-\sqrt{2x+1})(\sqrt{x+4}+\sqrt{2x+1})(\sqrt{3x+1}+\sqrt{x+7}) }{ (\sqrt{3x+1}-\sqrt{x+7})(\sqrt{3x+1}+\sqrt{x+7})(\sqrt{x+4}+\sqrt{2x+1} ) }=\]\[\frac{ (x+4-(2x+1))(\sqrt{3x+1}+\sqrt{x+7}) }{ (3x+1-(x+7))(\sqrt{x+4}+\sqrt{2x+1} ) }=\]\[\frac{ -(x-3)(\sqrt{3x+1}+\sqrt{x+7}) }{ 2(x-3)(\sqrt{x+4}+\sqrt{2x+1} ) }=\]\[-\frac{ 1 }{ 2 }\cdot \frac{ (\sqrt{3x+1}+\sqrt{x+7}) }{ (\sqrt{x+4}+\sqrt{2x+1} ) }\]Now let x be 3, to see what the limit will be: \[-\frac{ 1 }{ 2 }\cdot \frac{ 2\sqrt{10} }{ 2\sqrt{7} }=-\frac{ 1 }{14 }\sqrt{70}\]

Answer 9

ahh..., there will be cancellation...

thanks a lot... :)