# Limit question

• Nov 3rd 2006, 09:21 PM
killasnake
Limit question
Sorry, I have no idea how to solve this limit problem

Quote:

Use a CAS (Computer Algebra System) to evaluate the following limit:
http://img.photobucket.com/albums/v6...25750img12.gif
I got this answer but it's wrong

cos(0)-sec(0)^2)/2
• Nov 4th 2006, 12:01 AM
Soltras
Well if you use something like Mathematica (software) it should give you 1/24 as the answer.

To solve it analytically, the easiest way is to use L'Hopital's Rule.

$\lim_{x \to 0} \frac{\cos x - 1 + \frac{x^2}{2}}{x^4}$

If you substitute 0 in for $x$, then you get $\frac{0}{0}$ which is criterion for using L'Hopitale.

Derive the numerator and denominator with respect to $x$.
The limit becomes:

$\lim_{x \to 0} \frac{-\sin x + x}{4x^3}$

Once again, direct substitution will yield $\frac{0}{0}$, so L'Hopitale can be applied again, giving the limit:

$\lim_{x \to 0} \frac{-\cos x + 1}{12x^2}$

Again, direct substitution gives $\frac{0}{0}$. Do L'Hopitale yet again to get the limit:

$\lim_{x \to 0} \frac{\sin x}{24x}$

You may recognize this form as $\frac{1}{24} \cdot \left( \lim_{x \to 0} \frac{\sin x}{x} \right) = \frac{1}{24} \cdot (1) = \frac{1}{24}$.

(if you didn't recognize the form, applying L'Hopitale one last time would give 1/24 trivially).
Hope that helps.
• Nov 4th 2006, 01:47 AM
CaptainBlack
Quote:

Originally Posted by killasnake
Sorry, I have no idea how to solve this limit problem

I got this answer but it's wrong

cos(0)-sec(0)^2)/2

I dont know what you are doing with your CAS but see attachment.

RonL