Looking for something else?

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

- anonymous

can i get some help with limits?

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.

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

- anonymous

can i get some help with limits?

- jamiebookeater

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

Get this expert

answer on brainly

SEE EXPERT ANSWER

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

and **thousands** of other questions

- KingGeorge

Almost certainly. What kind of limits do you need help with?

- anonymous

i have to show that f(x) has a removable discontiunty at x=2 and for what value of f(2) f(x) countionous. the equation is \[(2x^2+3x-14)/(x-2)\]

- anonymous

check to see if 2 makes the numerator zero. if it does factor and cancel

Looking for something else?

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

- anonymous

you get
\[(x-2)(2x+7)\] canceling gives you
\[2x+7\] replace x by 2 and get 11

- anonymous

i don't really get continuity at all i guess

- anonymous

ok the algebra is easy enough yes? so now the question is, what is the difference between
\[f(x)=\frac{2x^2+3x-14}{x-2}\] and
\[g(x)=2x+7\]which is just a line

- KingGeorge

You can only factor and cancel here because the denominator is equal to 0 at x=2. It's a little trick that follows from some theorems learned in pre-calc or calc 1.
To show it has removable discontinuity at x=2, you only have to say that the only place the function is undefined, is at x=2 (obvious since you can't divide by 0), and show that a limit exists at x=2.
To show the limit exists, you can check if the numerator, as well as the denominator is equal to 0 at x=2. If this is true, factor, cancel, and solve. Remember, this only works if the numerator AND the denominator are equal to 0.

- anonymous

and the answer is that they are identical, except for the fact that f is not defined at 2, but g is

- anonymous

so the graph of f looks just like the graph of g, a line with slope 2 and y intercept 7, except that at (2,11) where the function ought to be 11 there is a hole. other than that, they are identical

- anonymous

meaning that if you define f to be 11 when x = 2 now it is in fact identical to the line
\[y=2x+7\] and you have "removed" the discontinuity

- anonymous

sorry it took me so long respond, this online work is being very slow... anyway say i have 2 equations \[(4x^3+24x^2+7x+42)/(x+6) and (2x^2+6x+a)\] the first formula has an x<-6 and the 2nd has x

Looking for something else?

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