00:01
Hi there, so for the limit a, we have that this is the limit, when x tends to 5 to the right, of x squared and this minus 25, this divided by x squared minus six times x and this plus 5.
00:32
So we need to solve this limit.
00:38
Now, the first thing that we can do is to try to simplify this.
00:47
So, for example, the one in the numerator, we can write it as x plus 5, and this times x minus 5.
00:58
And then the one in the denominator, we can write it as well as a product.
01:08
So in this case, that could be minus 5 and minus 1.
01:16
So we can cancel this with this one right here.
01:24
So now if we substitute the value of 5 in there, there is already no determination.
01:31
So we will have that that is 10 divided by 5 minus 1, which is 4.
01:39
So we can simplify this as 5 divided by 2.
01:43
So that's a solution for par a of this problem.
01:49
Now for the limit b, we are told that this is the limit.
01:54
When x tends to 5 by the right, so there will be x to the square.
02:02
And this minus 306, this divided by x square minus six times x plus 5.
02:13
So let's do the same as before.
02:17
In this case, we can write that this is the limit when x tends to 5 by the right.
02:28
So that will be.
02:29
We can write this as x plus fifth times x minus six.
02:36
And then this divided by x minus 5, this times x plus, oh, sorry, minus 1.
03:03
Now with that said, we know that there is some determination in here, and we cannot cancel it for this.
03:11
But if we put 5 by the right, that means that is a value that is a little bit, greater than 5.
03:21
So with that we will have a value that is something divided by something very very little.
03:36
So that will give us since this is going to be greater than 0, so that will be positive infinity.
03:47
So that's the solution for part b of this problem.
03:53
Now for par c, we are asked about the limit when x tends to 1.
03:59
And this is x squared minus 1.
04:03
And this divided by x minus 1.
04:06
So we do the same as before.
04:07
So this is the limit with x stems to 1.
04:10
And we can write this as follows as s plus 1 times x minus 1.
04:16
And then this divided by x minus 1.
04:19
We can cancel this with this one here, and then we will obtain the limit when x times to 1 of x plus 1.
04:29
Now we can evaluate this at 1 because there is no indetermination in there, so that will give us 2.
04:37
So that's a solution for par c of this problem.
04:41
Now for power 10, we will have the limit when x tends to 0 of x plus 4 divided by x.
04:50
Now if we substitute 0 in there, we will have that this is, yes, if we substitute 0 in there, we will have that this is equal to 4 divided by 0.
05:12
So in order to see if this limit exists, what we are going to do is to do separately, the limit when x tends to 6.
05:20
0 by the right.
05:22
So that will be x plus 4 divided by x.
05:25
So in this case, we will have a value that is 0 plus plus 4 divided by 0 plus.
05:35
So that will be plus infinity.
05:43
And the limit when x tends to 0 by the left of 0 minus, sorry, it's plus 4.
05:57
Divided by x...