1. ## Limit

Here is a limit I received on a test the other day and I was wondering if there is another easier way to do it...and by the way you all have reformed me...I still love L'hopital's but I decided to be less lazy

$\displaystyle \lim_{x\to{0}}\frac{\sin(3x)(1-\cos(2x))}{e^{x^3}-1}$

Imputting the power series we get

$\displaystyle \lim_{x\to{0}}\frac{\bigg(3x-\frac{(3x)^3}{3!}+\frac{(3x)^5}{5!}+\text{...}\big g)\bigg(1-\bigg[1+\frac{(2x)^2}{2!}+\frac{(2x)^4}{4!}+\text{...}\b igg]\bigg)}{\bigg[1+x^3+\frac{x^6}{2!}+...\bigg]-1}$

Cancelling terms and changing the signs on the top right parenthases we get

$\displaystyle \lim_{x\to{0}}\frac{\bigg(3x-\frac{(3x)^3}{3!}+\frac{(3x)^5}{5!}+...\bigg)\bigg (\frac{(2x)^2}{2!}-\frac{(2x)^4}{4!}-...\bigg)}{x^3+\frac{x^6}{2!}+...}$

I dont know how to display this in correct notation (suggestions would be helpful!) But what is concise mathematical notation to say that since we are going to 0 the only terms that affect the limit are the lowest exponent terms? I know you can use "Big Oh" notation in algorithim discreet mathematics but I have only seen that as things get sufficiently big not small...but anyways because of this fact this limit is equivalent to

$\displaystyle \lim_{x\to{0}}\frac{(3x)(\frac{(2x)^2}{2!})}{x^3}= \lim_{x\to{0}}6=6$

Is this corect and/or is there an easier way? Thanks in advance!

2. Originally Posted by Mathstud28
Here is a limit I received on a test the other day and I was wondering if there is another easier way to do it...and by the way you all have reformed me...I still love L'hopital's but I decided to be less lazy

$\displaystyle \lim_{x\to{0}}\frac{\sin(3x)(1-\cos(2x))}{e^{x^3}-1}$

Imputting the power series we get

$\displaystyle \lim_{x\to{0}}\frac{\bigg(3x-\frac{(3x)^3}{3!}+\frac{(3x)^5}{5!}+ \text{...}\bigg)\bigg(1-\bigg[1+\frac{(2x)^2}{2!}+\frac{(2x)^4}{4!}+\text{...}\b igg]\bigg)}{\bigg[1+x^3+\frac{x^6}{2!}+...\bigg]-1}$

Cancelling terms and changing the signs on the top right parenthases we get

$\displaystyle \lim_{x\to{0}}\frac{\bigg(3x-\frac{(3x)^3}{3!}+\frac{(3x)^5}{5!}+...\bigg)\bigg (\frac{(2x)^2}{2!}-\frac{(2x)^4}{4!}-...\bigg)}{x^3+\frac{x^6}{2!}+...}$

I dont know how to display this in correct notation (suggestions would be helpful!) But what is concise mathematical notation to say that since we are going to 0 the only terms that affect the limit are the lowest exponent terms? I know you can use "Big Oh" notation in algorithim discreet mathematics but I have only seen that as things get sufficiently big not small...but anyways because of this fact this limit is equivalent to

$\displaystyle \lim_{x\to{0}}\frac{(3x)(\frac{(2x)^2}{2!})}{x^3}= \lim_{x\to{0}}6=6$

Is this corect and/or is there an easier way? Thanks in advance!
Once here:

$\displaystyle \lim_{x\to{0}}\frac{\bigg(3x-\frac{(3x)^3}{3!}+\frac{(3x)^5}{5!} {\color{red}-} ...\bigg)\bigg(\frac{(2x)^2}{2!}-\frac{(2x)^4}{4!} {\color{red}+} ...\bigg)}{x^3+\frac{x^6}{2!}+...}$

my next step would be to divide by the common factor x^3 = (x)(x^2):

$\displaystyle \lim_{x\to{0}}\frac{\bigg(3-\frac{(3)^3 x^2}{3!}+\frac{(3^5) x^4}{5!} {\color{red}-} ...\bigg)\bigg(\frac{2^2}{2!}-\frac{(2^4) x^2}{4!} {\color{red}+} ...\bigg)}{1 +\frac{x^3}{2!}+...}$

and then take the limit.