00:01
In this question, we are asked to calculate the given limit.
00:03
First of all, note that if you plug in x equals 1 directly, you're going to get 0 in the denominator of the first fraction and 0 in the denominator of the second fraction.
00:16
So we can't really do that.
00:18
What we can do instead is rewrite the expression inside the limit under the common denominator.
00:25
To do that, we need to multiply the first fraction by lnx and the second fraction by x minus 1.
00:32
And what we are going to get is x times ln x minus x minus 1 divided by x minus 1 times ln x let's simplify that expression so this equals to the limit of x lnx minus x plus 1 divided by x minus 1 multiplied by ln x now if we plug in x equals 1 we are going to get 0 in the numerator and and also zero in the denominator.
01:19
So that's going to be an indeterminate form, 0 over 0.
01:27
And this also means that we can apply the lopetals rule to the limit.
01:34
By the lopetals rule, the original limit equals to the limit of the ratio of the derivatives, which means we need to differentiate the numerator and the denominator separately.
01:47
The derivative of the numerator is going to be lnx plus x multiplied by 1 or x...