Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

find the limit as x approaches 9 : WILL GIVE EQUATION IN RESPONSE

See more answers at
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.

Get this expert

answer on brainly


Get your free account and access expert answers to this and thousands of other questions

\[\lim_{x \rightarrow 9}(\frac{ 1 }{ \sqrt{x}-3 } - \frac{ 6 }{ x-9 })\]
do you know l hospital's rule?
please don't use lhopitals or anything im in basic calculus and haven't learned it yet. Suposed to be able to algebriaically change it to solve. I know the answer is suposed to be 1/6 i just don't know how to get there

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

ok by basics , rationalise first fraction
multiply by \(\sqrt{x}+3\) on numerator and denominator
so the left turns into right?
ugh didn't get the equation in
\[\frac{ \sqrt{x}+3 }{ x-3 }\]
try again ?
is that not right after multiplying by \[\sqrt{x}\]+3?
so i now have \[\frac{ \sqrt{x}+3 }{ x-9 } - \frac{ 6 }{ x-9}\]
ok now add them !! note denominator is same
well that gives me zero... and i know the answer is suposed to be 1/6
try again?
right, nevermind. thanks!
no i still get zero
\[\frac{ \sqrt{x}+3 }{ x-9 }- \frac{ 6 }{ x-9 }\]
\[\frac{ \sqrt{x}-3 }{ x-9 }\] and if i plug in 9 it will be 0 over 0
please dont jump into putting vale ! we are not over yet!
ok, what do i do if i dont put the value in then
because your calculations give me \[\lim_{x \rightarrow 9}\frac{ \sqrt{x}-3 }{ x-9 }\]
do i split the bottom up? into its factors?
found it out! thanks!!
yw : )

Not the answer you are looking for?

Search for more explanations.

Ask your own question