Fundamental Limits

[varun.amin]

Hey,

I was wondering if I could get some help on the question attached? The solutions have given the answer as 0.5. Thank you in advance!

[TrueTears]

\begin{align*}
\lim_{x \rightarrow 0} \frac{1-\cos(x)}{x^2} & = \lim_{x \rightarrow 0} \frac{1-\left[1-\frac{x^2}{2} + \mathcal{O}(x^4) \right]}{x^2} \\
& = \frac{1}{2} + \mathcal{O}(x^2) \\
& = \frac{1}{2}
\end{align*}

[varun.amin]

\begin{align*}
\lim_{x \rightarrow 0} \frac{1-\cos(x)}{x^2} & = \lim_{x \rightarrow 0} \frac{1-\left[1-\frac{x^2}{2} + \mathcal{O}(x^4) \right]}{x^2} \\
& = \frac{1}{2} + \mathcal{O}(x^2) \\
& = \frac{1}{2}
\end{align*}

Hi TrueTears, thanks for replying! I don't quite understand the solution, could you please break it down for me?

[RuiAce]

\begin{align*}
\lim_{x \rightarrow 0} \frac{1-\cos(x)}{x^2} & = \lim_{x \rightarrow 0} \frac{1-\left[1-\frac{x^2}{2} + \mathcal{O}(x^4) \right]}{x^2} \\
& = \frac{1}{2} + \mathcal{O}(x^2) \\
& = \frac{1}{2}
\end{align*}

O-notation is not included in the HSC at all.

Hey,

I was wondering if I could get some help on the question attached? The solutions have given the answer as 0.5. Thank you in advance!

[TrueTears]

For the first equality, a Taylor series expansion is taken to the fourth order, second equality is just algebra, last equality holds by definition of big-O:
\begin{align*}
f(x) = \mathcal{O}(x^2) \iff |f(x)| \le c|x^2|
\end{align*}
for some constant . Taking limits as and applying Squeeze theorem yields .

[varun.amin]

O-notation is not included in the HSC at all.

Thank you