00:02
In this problem, we are provided with the following values.
00:06
We have the x values as 3.
00:12
3 .9, 3 .99, 3 .999, 3 .999, 5, 4 .1, 4 .01, 4 .000, 0 .00001.
00:26
And the corresponding values of f of x are 6, 7 .92, 7 .99 -2, 7 .99, 2.
00:37
7 .000, double 9, double 9, double 9, double 9, double 9 .002.
00:43
And for x equals to 5, we have 11.
00:47
Next, we have 9 .07, 9 .0000007.
00:53
9 .000 -tru -0 -0 -3 -0 -0 .7.
00:58
We are asked to find out the value of limit x tends to 4 f of x.
01:04
So here, if we consider the left -hand limit, we see that we obtain the value to be 8, since it can be seen that as we are approaching the value of x equals to 4, the value of f -fx tends to 8.
01:21
And if we consider the right -hand limit, that is x tends to 4 positive, we see that as we are approaching x equals to four the value of f is approaching nine so from this we can see that the left hand limit and the right hand limit are not equal which means that the limit does not exist d n e this is the required answer for the first question in the second question we are asked to find the value of limit x tends to negative 1 x squared plus 3 times x minus 4 the whole divided by x squared minus 16.
02:05
So let us substitute x as negative 1.
02:07
We have negative 1 the whole squared plus 3 times negative 1 minus 4.
02:12
The whole divided by negative 1 the whole squared minus 16.
02:16
Simplifying this we get 1 minus 3 minus 4.
02:20
The whole divided by 1 minus 16.
02:23
Further simplification gives negative 6 over 9.
02:25
Negative 15 which further simplifies to 2 over 5 so this is the required answer of sub part 2 in sub part 3 we are asked to evaluate limit x tends to 5 square root of 6x minus 4 so let us substitute the value of excess 5 we have 6 times 5 minus 4 6 times 5 is 30 so we get 30 minus 4 which equals to 26.
02:57
So therefore the required limit is square root of 26.
03:02
In subpart 4, we are asked to find out limit x tends to 2 x minus g of x over f of x and we are given that limit x tends to 2 f of x equals to negative 3 and limit x tends to 2 g of x equals to 7 so taking the limit we have limit x tends to 2 g of x minus limit x tends to 2 g of x the whole divided by limit x tends to 2 f of x so now let us substitute the values we get 2 minus 7 the whole divided by negative 3 so we obtain negative 5 over negative 3 which simplifies to 5 over 3.
03:56
So this is a required answer for subpart 4.
04:01
In subpart 5 we are asked to find out the points of discontinuity of 5x over x squared plus 16.
04:11
So here it can be seen that this function is defined everywhere.
04:23
That is, it exists for all values of x...