anonymous
  • anonymous
solve limx4 (sqrt 8-x)-2/(sqrt 5-x)-1
Mathematics
  • 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.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
can you please show me step by step thanks
angela210793
  • angela210793
lim x-->4 F(x)=F(4)=sqr (8-4)-2/sqr(5-4)-1=0/0 indefinite form
angela210793
  • angela210793
so we can use Hopital's rule

Looking for something else?

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

More answers

angela210793
  • angela210793
http://en.wikipedia.org/wiki/L'H%C3%B4pital's_rule
angela210793
  • angela210793
lim x-->4 [(sqr8-x)-2]'/[(sqr5-x)-1]'
angela210793
  • angela210793
Do u know derivatives?
anonymous
  • anonymous
i got the denomiator 0 and the numerator zero as well
anonymous
  • anonymous
not yet im currently taking calc
angela210793
  • angela210793
yea u can use Hopital's rule when u get 0/0 or infinite/infinite
anonymous
  • anonymous
ok thanks
angela210793
  • angela210793
[(sqr8-x)-2]'=1/2sqr(8-x) and [(sqr5-x)-1]'=1/2sqr(5-x) so we'll have 1/2sqr(8-x) 2sqr(5-x) lim x-->4 -------------=lim x-->4------------=1/2 1/2sqr(5-x) 2sqr(8-x)
angela210793
  • angela210793
Got it?
anonymous
  • anonymous
yes!! now i understand completely thank you soooo much..Im gonna refer to this as much as i can remeber
angela210793
  • angela210793
Happy i helped :)

Looking for something else?

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