00:01
In this portion, we need to obtain the limits of the given piecewise functions.
00:07
The first given piecewise function is h of x is equal to 1 plus x squared if x is less than equal to 0 and x squared minus x plus 1 if x is greater than 0.
00:29
Now for this function, the first limit that we need to obtain our limit x approaches to 0 negative, that is left -hand limit of h of x at x is equal to 0.
00:44
Now, when x is less than 0, the value of function is 1 plus x squared, that is limit x approaches to 0 negative 1 plus x square.
00:58
Substituting the limit we get 1 plus 0 square is equal to 1.
01:03
Further, the second limit is limit x approaches to 0 positive h of x.
01:11
That means the right hand limit of h of x x x is equal to 0, which is limit x approaches to 0 positive x square minus x plus 1.
01:23
Because h of x is x minus x plus 1 when x is greater than 0.
01:29
This gives 0 minus 0 plus 1 unsucceed in the limit, which is equal to 1.
01:36
Now this shows that the left -hand limit of h of x is equal to the right -hand limit of h of x.
01:48
Therefore, the limit exists...
