Continuity at a Point

For a function,

 $f\left(x\right)$  to be continuous at a point  $x=c$  , which of the following conditions must be met:

i.  $\lim_{x\rightarrow c}f\left(x\right)$ must exist

ii.  $f\left(c\right)$ must be defined  

iii.  $\lim_{x\rightarrow c}f\left(x\right)=f\left(c\right)$  

• i only

• i and ii only

• iii only

• i, ii, and iii

True or false: The piece-wise function,  $f\left(x\right)$ , is continuous at  $x=2$ :  $f\left(x\right)=\sqrt{x+7},\ x\le2$  

 $f\left(x\right)=x^2-1,\ x>2$  

• True

• False

True or false: The piece-wise function,  $f\left(x\right)$ , is continuous at  $x=\frac{\pi}{4}$ :  $f\left(x\right)=\cos x,\ x<\frac{\pi}{4}$  

 $f\left(x\right)=\sin x,\ x>\frac{\pi}{4}$  

• True

• False

True or false: The piece-wise function,  $f\left(x\right)$ , is continuous at  $x=0$ :  $f\left(x\right)=\frac{\sin x}{x},\ x<0$  

 $f\left(x\right)=\cos x,\ \ x\ge0$  

• True

• False

Which function below has a non-removable, infinite discontinuity at  $x=-1$  ?

•  $y=\frac{1}{x^2+1}$  

•  $y=\frac{x}{x+1}$  

•  $y=\frac{\left|x+1\right|}{x+1}$  

•  $y=\frac{x^2-1}{x+1}$  

Which function below has a removable discontinuity at  $x=-1$  ?

•  $y=\frac{1}{x^2+1}$  

•  $y=\frac{x}{x+1}$  

•  $y=\frac{\left|x+1\right|}{x+1}$  

•  $y=\frac{x^2-1}{x+1}$  

Which type of discontinuity is shown in the graph?

• removable

• non-removable, infinite

• non-removable, jump

• it is continuous everywhere

Identify each x-value for which there is a discontinuity if:  $f\left(x\right)=-x,\ \ x<-1$  and

  $f\left(x\right)=-\frac{1}{x},\ -1\le x\le1$ , and

  $f\left(x\right)=\sqrt{x},\ \ x>1$   

i. discontinuity at  $x=-1$  

ii. discontinuity at  $x=0$  

iii. discontinuity at  $x=1$  

• i only

• ii only

• ii and iii only

• iii only

Find the value of 'a', that makes the piece-wise function continuous if:

 $f\left(x\right)=2x-3,\ \ x<2$  

and 

 $f\left(x\right)=x^2+ax-2,\ \ x\ge2$  

•  $a=-\frac{1}{2}$  

•  $a=\frac{1}{2}$  

•  $a=-\frac{3}{2}$  

•  $a=\frac{3}{2}$