# Given h(x)={-3x-3 if x<-1; 1-X^2 if -1<x<2; -4 is x>= 2 A) limh(x) B). lim (X) x-->>-1+  x-->-1-

mlehuzzah | Student, Graduate | (Level 1) Associate Educator

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`lim_(x->-1^+)`

means the limit as x approaches -1 from the right (that is, start "above" or greater than x, and move towards x)

So we want to look at numbers just slightly greater than -1

For these numbers (in this case, -1<x<2), h(x) is given by 1-x^2

` 1-x^2`  is a polynomial, so it is continuous, so the limit can be obtained by plugging in x=-1: `1-(-1)^2=0 `

Thus:

`lim_(x->-1^+) h(x)=0`

`lim_(x->-1^-)`  means the limit as x approaches -1 from the left (that is, start "below" or less than x, and move towards x)

So we want to look at numbers just slightly less than -1

For these numbers (x<-1), h(x) is given by -3x-3

`-3x-3`  is continous (again it is a polynomial), so you can find the limit by plugging in x=-1: `-3(-1)-3=0`

so the limit is 0:

` ` `lim_(x->-1^-)h(x)=0`

Note: if both the left handed and right handed limits exist and are the same, then the limit exists, and:

`lim_(x->-1)h(x)=0`

PS: It might be helpful to visualize these on a graph:

Start to the left of x=-1.  Travel down the line towards x=-1, and you approach y=0.  (This is the left-handed limit)

Start to the right of x=-1.  Travel over the parabola towards x=-1, and you approach y=0.  (This is the right-handed limit)