Limits help (kind of forgot...) Click here to see function

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 our expert's

answer on brainly

SEE EXPERT ANSWER

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

A community for students.

Limits help (kind of forgot...) Click here to see function

Mathematics
See more answers at brainly.com
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

SEE EXPERT ANSWER

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

\[ \huge \lim_{x \rightarrow 2} \frac{(x-3)(x+2)}{(x-2)}.\]
How can I simplify the denominator so that it doesnt give me a 0..?
numerator is not zero, so go fish

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

directly put x=2
i.e. no limit
? Wouldn't it be a a denominator of 0, which is a "no-no"?
if you get a zero in the denominator, but not a zero in the numerator, then there is no limit only when you get \(\frac{0}{0}\) can you continue if you get a zero in the denominator
no
No, We haven't learned about that yet
l'hopital works for \(\frac{0}{0}\) in any case you didn't get there yet i am sure
So, @satellite73 would the limit be DNE ?
Because my teacher always says to only substitute directly when the denominator is not equal to zero...
not applicable here if you have a rational funciton, and you want to take the limit as x goes to some number, the first step is to plug in the number if you get a number back, that is your answer if you get \(\frac{a}{0}\) where \(a\neq 0\) there is no limit if you get \(\frac{0}{0}\) there is more work to be done factor and cancel but in this case you get \(\frac{-4}{0}\) so forget it
That makes a lot of sense @satellite73! Thanks so much!!

Not the answer you are looking for?

Search for more explanations.

Ask your own question