# Consider the piecewise function f(x) = (4x^2 -3)/(3x^2-15) , x less than -2 (4-x^2)^(1/2) , -2 less than = to x less than= to 2...

Consider the piecewise function

f(x) = (4x^2 -3)/(3x^2-15) , x less than -2

(4-x^2)^(1/2) , -2 less than = to x less than= to 2

5 , x greater than 2

a) Draw a neat sketch of its graph.

b) use algebra to find lim x--> -infinity f(x)

c) find the following ;

lim x--> (-2)^ - f(x) = ?

lim x--> (-2)^+ f(x)= 0

f(-2)= 0

f(4)= 5

And list the values of c for which f(x) is not continuos :

*print*Print*list*Cite

### 1 Answer

The function f(x) is defined over three regions.

f(x) =

`(4x^2-3)/(3x^2-15)` `xlt-2`

`sqrt((4-x^2))` `-2lt=xlt=2`

`5` ` xgt2`

When x goes to `-oo` , the relevant function is the first one, when `xlt-2`

`lim_(x-gt-oo)(4x^2-3)/(3x^2-15)`

you can't calculate the limit straight away, but you can divide both numerator and denominator from x^2 and modify this expresssion.

`lim_(x-gt-oo)((4x^2-3)/x^2)/((3x^2-15)/x^2)`

`lim_(x-gt-oo)(4-3/x^2)/(3-15/x^2)`

Now if we take the limit,

`lim_(x-gt-oo)(4-3/(-oo)^2)/(3-15/(-oo)^2) = (4-0)/(3-0)`

Therefore,

`lim_(x-gt-oo)f(x) = 4/3`

`lim_(x-gt-2^-)f(x)`

This is the limit when x is approaching to -2 from the negative infinity side. Therefore still the function is the first one,

`lim_(x-gt-2^-)f(x) =` `lim_(x-gt-2^-)[(4x^2-3)/(3x^2-15)] = (4*(-2)^2-3)/(3*(-2)^2-15)`

` ` `lim_(x-gt-2^-)f(x) =13/(-3) = -13/3`

`lim_(x-gt-2^+)f(x)`

This is the limit when x is approaching to -2 from the positive infinity side(or decreasing to -2). Therefore now the function is the second one.

`lim_(x-gt-2^+)f(x) = lim_(x-gt-2^+)sqrt(4-x^2)`

Therefore,

`lim_(x-gt-2^+)f(x) = 0`

If you have noticed,

`lim_(x-gt-2^-)f(x) != lim_(x-gt-2^+)f(x)`

So at x =-2 , f(x) has a jump discontiuity.

The other discontinuity is at x=2.

So there are discontinuities at x =-2 and x =2.

`f(-2) = sqrt(4-(-4)^2) = 0`

`f(4) = 5` (since 4>2)