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.
Follow Math Help Forum on Facebook and Google+
The answer is 15 as you determined. Whoever wrote the answer key wasn't paying attention.
Many Thanks.
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
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