Continuous

**Petrus** · February 8th 2013, 11:29 AM

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).

**HallsofIvy** · February 8th 2013, 11:40 AM

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)?

**Petrus** · February 8th 2013, 11:53 AM

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)

**Plato** · February 8th 2013, 02:04 PM

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

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

**Petrus** · February 8th 2013, 02:19 PM

If i make it to a equation then i get c=1 so?

**Prove It** · February 8th 2013, 02:28 PM

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?

**Plato** · February 8th 2013, 02:43 PM

Originally Posted by Petrus

If i make it to a equation then i get c=1 so?

Surly you get $c=\frac{2}{3}~?$

**Petrus** · February 8th 2013, 02:54 PM

Sorry My bad, accident did write a wrong number on equation:/ ty for all help,explain i got! Have a nice day

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