anonymous
  • anonymous
lim (4x-x^2)/(2-sqrtx) x->4 How is it 16?
Calculus1
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
factor and cancel, or multiply by the conjugate your choice
anonymous
  • anonymous
Alright satellite take it away :P .
Luigi0210
  • Luigi0210
Try multiply by the conjugate

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
I rationalized personally.
anonymous
  • anonymous
i pick factoring
Luigi0210
  • Luigi0210
Yea.. what Satellite said
anonymous
  • anonymous
damn typo \[\frac{x(4-x)}{2-\sqrt{x}}=\frac{x(2-\sqrt{x})(2+\sqrt{x})}{2-\sqrt{x}}\]
anonymous
  • anonymous
Was about to say :P .
anonymous
  • anonymous
cancel, replace \(x\) by \(16\)
anonymous
  • anonymous
Replace x with 4 not 16 :P .
anonymous
  • anonymous
Oh alright thanks a lot! Sometimes I get stuck on little things.
anonymous
  • anonymous
you can do what @Luigi0210 said too try it and see multiply top and bottom by \(2+\sqrt{x}\) leave the numerator in factored form, cancel the \(4-x\) top and bottom
anonymous
  • anonymous
oh right, replace \(x\) by \(4\) not \(16\)
anonymous
  • anonymous
@Dido525 keeping me honest here ...
anonymous
  • anonymous
I try :) .
anonymous
  • anonymous
L'hospital's rule would work here too I think.

Looking for something else?

Not the answer you are looking for? Search for more explanations.