Sorry, I have no idea how to solve this limit problem
I got this answer but it's wrongUse a CAS (Computer Algebra System) to evaluate the following limit:
cos(0)-sec(0)^2)/2
Sorry, I have no idea how to solve this limit problem
I got this answer but it's wrongUse a CAS (Computer Algebra System) to evaluate the following limit:
cos(0)-sec(0)^2)/2
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.
Here's what you start with.
$\displaystyle \lim_{x \to 0} \frac{\cos x - 1 + \frac{x^2}{2}}{x^4}$
If you substitute 0 in for $\displaystyle x$, then you get $\displaystyle \frac{0}{0}$ which is criterion for using L'Hopitale.
Derive the numerator and denominator with respect to $\displaystyle x$.
The limit becomes:
$\displaystyle \lim_{x \to 0} \frac{-\sin x + x}{4x^3}$
Once again, direct substitution will yield $\displaystyle \frac{0}{0}$, so L'Hopitale can be applied again, giving the limit:
$\displaystyle \lim_{x \to 0} \frac{-\cos x + 1}{12x^2}$
Again, direct substitution gives $\displaystyle \frac{0}{0}$. Do L'Hopitale yet again to get the limit:
$\displaystyle \lim_{x \to 0} \frac{\sin x}{24x}$
You may recognize this form as $\displaystyle \frac{1}{24} \cdot \left( \lim_{x \to 0} \frac{\sin x}{x} \right) = \frac{1}{24} \cdot (1) = \frac{1}{24}$.
(if you didn't recognize the form, applying L'Hopitale one last time would give 1/24 trivially).
Hope that helps.