00:01
We're going to evaluate the following limits.
00:04
In part 1, the limit when x goes to 1 of 2x minus x to the 4th to the 1 half minus x to the 1 third, and all that divided by 1 minus x to the 3 fourth.
00:14
Part 2, the limit when x goes to plus infinity of 5x squared minus 3x over 7x squared plus 1.
00:26
In part 3, limit when x goes to infinity of x minus square root of x plus x squared.
00:33
And in part 4, limit when x goes to plus infinity of the square root of x plus 2 divided by the square root of x plus 1.
00:42
So let's start with limit in part 1.
00:45
So remember, the first thing we got to do when x is converted into a value is to try to evaluate the expression to see if that can be done.
00:55
In this case, when we evaluate the expression at x equal 1, we get 2 minus 1 to the 1 half, because we have 2 times 1, and 1 to the 4th.
01:10
So we get 2 minus 1 to the 1 half minus 1 to the 1 third, all that divided by 1 minus 1 to the 3 fourth.
01:19
And so we get 2 minus 1 is 1, 1 to the 1 half minus 1 to the 1 third divided by 1 minus 1 to the 3 fourth, and you get here 1 minus 1 and 1 minus 1, both in the numerator and denominator, so it's 0 over 0, which is not a number.
01:42
That is, when we do the evaluation we get 0 divided by 0, which is not a number, that is, that cannot be the limit.
01:50
In fact, when we get something like this, we say that the limit is undetermined of the type or of the form, let's say better, of the form 0 divided by 0.
02:18
That is, both numerator and denominator separately, both converges to 0 when x converges to 1.
02:28
That's why we say it's undetermined of the form 0 divided by 0.
02:33
When we have such a type of such a form limit, we can apply l 'hopital's rule.
02:43
L 'hopital's rule, i guess with t is double, l 'hopital's rule, in which we find the derivative of the numerator and the derivative of the denominator separately.
03:03
So we can say that the limit when x goes to 1 of the original expression to x minus x to the fourth to the one -half minus x to the one -third divided by 1 minus x to the 3 -fourth is exactly equal the following l 'hopita's rule to the limit when x goes to 1 of the derivative with respect to x of the numerator to to x minus x to the fourth to the one -half minus x to the one -third, divided by the derivative of the denominator.
03:51
So it's very important to remark we are calculating the derivative separately.
03:57
And lobito's rule says the limit of the expression, which is undetermined of the form 0 divided by 0, that is, it's important to verify that before applying the rule, because the rule only applies for that type of limit.
04:11
And all the types but in particular in this for this type of limit and so the rule says that the limit of the expression is equal to the limit of the derivative of the numerator divided by the derivative of the denominator separately that is we are not talking about the derivative of the quotient and so we calculate both derivatives to have this final expression so the derivative of the numerator is 1 half times 2x minus x to the fourth to the negative 1 half times, applying the shown rule, to 2 minus 4x cubed.
04:53
That's the derivative of the first term here, and then minus derivative of x to the one -third is one -third x to the negative two -thirds, and the derivative of the denominator, derivative of 1 is 0, so we get negative negative 3 fourth times x to the negative 1 fourth.
05:17
And we write this the following way, simplify a little bit, that's the limit when x goes to 1 of, so if we take the common factor out of the factor 2 minus 4x cubed, that factor out 2, going to simplify with the 2 here in the denominator, and so the expression in the numerator becomes 2x minus x to the fourth to the negative one -half times 1 minus 2x cubed after taking two common factor out and simplify it with the two in the denominator minus 1 over 3x to the two -thirds because we have the negative exponent in the numerator, we can put it in the denominator with positive exponent, over negative 3 fourths of x to the, sorry, in the denominator we do the same, that is, we can write this down as negative 3 over 4x to the 1 fourth.
06:23
And now we simplify a little bit more, we have also this exponent with negative sign so we can put it as limit when x goes to 1 of 1 over sorry not 1 but this polynomial here 1 minus 2x cubed over square root of 2x minus x to the fourth minus 1 over 3x to the two -thirds and all that divided by negative 3 over 4x to the 1 fourth and doing a final simplification.
07:05
This is the limit when x goes to 1.
07:08
We write down the final expression when we simplify more to the fraction operations and all that or maybe we don't do the fraction operation but we do the division here maybe.
07:21
Bit so we get 1 over 3x to the 2 thirds plus 2x cubed minus 1 divided by the square root of 2x plus minus or x to the fourth all that multiplied by 4x to the 1 fourth divided by 3 that's the expression we get and now we proved by evaluating now the expression we got at the end here at x equal 1 because x is converging to 1 so we try to evaluate now the expressions we get 1 over 3 she's putting x equal 1 here plus 2 times 1 cube minus 1 is 2 minus 1 is 1 over square root of 2 times 1 is 2 minus 1 to the 4th is 2 minus 1 all that that multiplied by 4 times 1 to the 1 fourth is 4 over 3.
08:24
So we get 1 third plus 1 over 1 times 4 thirds, and that is 4 thirds cubed, sorry, squared, because this expression here is 4 thirds.
08:43
Multiply by 4 thirds, 4 thirds squared, so it's 16 over 9.
08:48
So the limit we were calculating when x converges to 1 of 2x minus x to the 4th to the 1 half minus x to the 1 3rd over 1 minus x to the 3 4th is equal to 16 over 9.
09:11
That's part 1.
09:15
Good.
09:16
Now we do part 2.
09:17
That is, we calculate the limit when x goes to plus infinity of 5x squared minus 3x divided by 7x squared plus 1...