# Thread: continuity

1. ## continuity

a is a real number.
f:R--->R, f(x)= 2x+a,x>=3
16-ax^2, x<3

If this function is continious find lim f(x)
x-->a?

I found a=1
6+a=16-9a
a=1
when x--->1
16-x^2=15
and found the limit as 15.But book says 16-a.

2. ## Re: continuity

The answer is 15 as you determined.

Whoever wrote the answer key wasn't paying attention.

Many Thanks.

4. ## Re: continuity

Originally Posted by kastamonu
a is a real number.
f:R--->R, f(x)= 2x+a,x>=3
16-ax^2, x<3
If this function is continious find lim f(x) .
$f(x)=\begin{cases}2x+a & x\ge 3 \\16-ax^2 & x<3\end{cases}$____$\begin{cases}\displaystyle{\lim _{x \to {3^ + }}}f(x) = 6 + a & \\\displaystyle{\lim _{x \to {3^ - }}}f(x) = 16-9a & \end{cases}$

a=?

a=1

6. ## Re: continuity

Originally Posted by kastamonu
a is a real number.
f:R--->R, f(x)= 2x+a,x>=3
16-ax^2, x<3

If this function is continious find lim f(x)
x-->a?

I found a=1

6+a=16-9a
a=1
when x--->1
16-x^2=15
and found the limit as 15.But book says 16-a.
Since a= 1, 16- a is 15