jokerzz said:Lim x-> 0 (e^xsin(x)-x-x^2)/(x^3 +xln(1-x))
I first put the Mclaurin series of e^x and sinx and then took x^3 common and then put the limit. My answer comes out to be 1/3. Can anyone confirm whether I've done it right?
Dick said:It would help if you showed a little more of the details, but there is an x^2 term in the denominator as well, isn't there? What did you do with that?
Unit said:I think your answer is a third too big.
jokerzz said:no there's no x^2. When you put x=0 at the end after substituting the mclaurin series of e^x and sinx, everything becomes 0, except -1/3! + 1/2 which is equal to 1/3
The limit without using L'Hospital's rule is a method of finding the limit of a function at a certain point without using derivatives. It involves algebraic manipulation and simplification of the function to determine the limit.
The limit without L'Hospital's rule should be used when the function is indeterminate at the given point, meaning that both the numerator and denominator approach zero or infinity.
To find the limit without using L'Hospital's rule, you will need to perform algebraic manipulations on the function to simplify it. This may involve factoring, cancelling out common terms, or using other algebraic techniques to rewrite the function in a simpler form.
Using the limit without L'Hospital's rule can help deepen your understanding of limits and how they relate to derivatives. It also allows you to solve for limits at points where L'Hospital's rule cannot be used, such as when the function is not differentiable at the given point.
While the limit without L'Hospital's rule can be a useful tool, it may not always be the most efficient method for finding limits. In some cases, using L'Hospital's rule or other techniques may be quicker and easier. Additionally, this method may not work for all types of functions, so it is important to consider other approaches as well.