# Continuous

• February 8th 2013, 11:29 AM
Petrus
Continuous
Hello i got problem with problem nr 41 . Ik how i shall think but that c making it difficoult for me. To se if its continuous is limit right side=limit left side=f(a).
• February 8th 2013, 11:40 AM
HallsofIvy
Re: Continuous
Quote:

Originally Posted by Petrus
Hello i got problem with problem nr 41 . Ik how i shall think but that c making it difficoult for me. To se if its continuous is limit right side=limit left side=f(a).

Very good! Not enough people know the basic definitions.
For x< 2, $f(x)= cx^2+ 2x$. What is $\lim_{x\to 2^-} f(x)$?
For x> 2, $f(x)= x^3- cx$. What is $\lim_{x\to 2^+} f(x)$?
What is f(2)?
• February 8th 2013, 11:53 AM
Petrus
Re: Continuous
Hello Hallsofivy!
The problem is idk how to calculate limit when i got c, well im kinda trying but does not work, im stuck there.
well this is what i think from left side we get 2(2+c)(negative way)
Right side 2(4-c) possitive way and f(2)=2(4-c)
• February 8th 2013, 02:04 PM
Plato
Re: Continuous
Quote:

Originally Posted by Petrus
Hello Hallsofivy!
The problem is idk how to calculate limit when i got c, well im kinda trying but does not work, im stuck there.
well this is what i think from left side we get 2(2+c)(negative way)
Right side 2(4-c) possitive way and f(2)=2(4-c)

You want $\lim _{x \to 2 ^ + } cx^2 + 2x = \lim _{x \to 2^ - } x^3 - cx$.

$\begin{gathered} \lim _{x \to 2^ - } cx^2 + 2x = 4c + 4 \hfill \\ \hfill \\ \lim _{x \to 2^ + } x^3 - cx = 8 - 2c \hfill \\ \end{gathered}$
• February 8th 2013, 02:19 PM
Petrus
Re: Continuous
If i make it to a equation then i get c=1 so?
• February 8th 2013, 02:28 PM
Prove It
Re: Continuous
Quote:

Originally Posted by Plato
You want $\lim _{x \to 2 ^ + } cx^2 + 2x = \lim _{x \to 2^ - } x^3 - cx$.

$\begin{gathered} \lim _{x \to 2^ - } cx^2 + 2x = 4c + 4 \hfill \\ \hfill \\ \lim _{x \to 2^ + } x^3 - cx = 8 - 2c \hfill \\ \end{gathered}$

Why didn't you substitute x = 2 into the LHS of your equation too?
• February 8th 2013, 02:43 PM
Plato
Re: Continuous
Quote:

Originally Posted by Petrus
If i make it to a equation then i get c=1 so?

Surly you get $c=\frac{2}{3}~?$
• February 8th 2013, 02:54 PM
Petrus
Re: Continuous
Sorry My bad, accident did write a wrong number on equation:/ ty for all help,explain i got! Have a nice day