Limits Evaluate the following limits using Taylor series.

Question

lim ₓ→₁ (x 1)/(ln x)

Accepted Answer

Hello. In this video, we are going to be evaluating the given limit by using a Taylor series. Now the limit that is given to us is the limit. As Z approaches 0, of E to the power of Z minus 1 divided by Z. Now, in order to approach this problem, we can approach this problem by recalling what the Taylor series for E to the power of Z is when it is centered at 0. For each of the power of Z when this Taylor series is centered at 0, this is equivalent to 1 plus Z. Plus Z squared divided by two factorial. Plus Z cubed divided by 3 factorial, plus an infinite amount of terms afterward. And so what we can go ahead and do in this case is we can approach this problem by building or expanding E to the Z and slowly transforming it into the function that is given in the limit. So what we've done so far is we stated the Taylor series expansion for E to Z, but what would happen if we were to apply a negative one to this infinite series? Well, if we were to subtract the entire series by 1, then the first term positive one will cancel out. So that is going to leave us with Z plus Z squared divided by 2 factorial, plus Z cubed divided by 3 factorial, plus an infinite amount of sums afterward. Then if we were to divide everything by Z. This will further simplify to be 1 plus Z divided by 2 factorial, plus Z squared divided by 3 factorial, plus an infinite amount of sums afterward. So what we've done here is we've found the Taylor series expansion for the function that's inside the limit. And what we can go ahead and do now is we can replace this expansion into the given limit in the problem. So we can rewrite the limit to be the limit as Z approaches 0 of the quantity 1 plus Z divided by 2 factorial, plus Z squared divided by 3 factorial, plus an infinite amount of sums afterward. Now, if we were to directly apply the limit in this part of the problem, notice that every term after the first constant term has some multiple of Z within the numerator. What that means is that when we directly apply the limit to every term, any term that has a Z in the numerator is going to become zero. So everything after the first term is going to become zero, and that means that the final value of the limit is going to be 1, and that is going to be the solution of the problem by using the Taylor series. And with that being said, the overall solution is going to be B. So I hope this video helps you in understanding how to approach this problem, and we will go ahead and see you all in the next video.

Limits Evaluate the following limits using Taylor series.
lim ₓ→₁ (x 1)/(ln x)

Key Concepts

Limits and Indeterminate Forms

Taylor Series Expansion

Logarithmic Function Behavior Near 1

Watch next