find the limit as h approaches 0 of h(3+(2/h^2))

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.

- clara1223

find the limit as h approaches 0 of h(3+(2/h^2))

- schrodinger

See more answers at brainly.com

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- clara1223

\[\lim_{h \rightarrow 0} h(3+\frac{ 2 }{ h ^{2} })\]

- amistre64

we can try some test points, say h=.0000001, and h=-.0000001
but what does your thoughts tell you initially?

- clara1223

Well I solved and got that the limit is 2, but that was wrong.

Looking for something else?

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

## More answers

- amistre64

your solution was flawed then :)
we have 2 terms: ab
if h < 0, a=-, b=+ ... ab = -
if h > 0, a=+,b=+ ... ab=+

- clara1223

I don't understand.

- amistre64

let a=h
let b=3+2/h^2
then multiply them together and yo have your setup.
when h is less than zero, what are our signs for a and b? what about their product?

- clara1223

if h is less than 0 then a would be negative and b would be positive, and their products would both be negative

- amistre64

ok, so we have a left side limit that is negative ...
now assume h is greater than zero ... what are the signs of a and b? and of their product?

- clara1223

if h is greater than 0 then a and b would both be positive and so would be their products.

- clara1223

oh, so since from the left it's negative and from the right it's positive, it doesnt exist?

- amistre64

so we have a right side limit that is positive
can a positive limit ever be equal to a negative limit?

- amistre64

correct, its actually -inf and +inf so yeah

- clara1223

yeah, I graphed it and it helped me to understand.

- amistre64

graphs do help if applicable :)

Looking for something else?

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