Thread: how to evaluate this limit at infinity

the answer is 2.42 but I don't know why to get there.

The first thing I would do is let u= 2.2/x. Then as x goes to infinity, u goes to 0 and the limit becomes
$\displaystyle \lim_{u\to 0}(2.2)^2{u^2}(1- cos(u))= \frac{1}{(2.2)^2}\lim_{u\to 0}\frac{1- cos(u)}{u^2}$.

Now apply L'hopital's rule.

Originally Posted by HallsofIvy

The first thing I would do is let u= 2.2/x. Then as x goes to infinity, u goes to 0 and the limit becomes
$\displaystyle \lim_{u\to 0}(2.2)^2{u^2}(1- cos(u))= \frac{1}{(2.2)^2}\lim_{u\to 0}\frac{1- cos(u)}{u^2}$.

Now apply L'hopital's rule.

Slight typo here, so to fix this up (and show an alternate solution):

$\displaystyle \begin{align*}\lim_{x \to \infty}\left(x^2-x^2\cos\left(\dfrac{2.2}{x}\right)\right) & = \lim_{u \to 0} \dfrac{(2.2)^2}{u^2}(1-\cos u) \\ & = (2.2)^2\lim_{u \to 0} \dfrac{1-\cos u}{u^2}\dfrac{1+\cos u}{1+\cos u} \\ & = 4.84\lim_{u \to 0} \dfrac{\sin^2 u}{u^2(1+\cos u)} \\ & = 4.84\left(\lim_{u \to 0}\dfrac{\sin u}{u}\right)^2\left(\lim_{u \to 0} \dfrac{1}{1+\cos u}\right) \\ & = 4.84(1)^2\left(\dfrac{1}{2}\right) = 2.42 \end{align*}$

The fourth line uses the limit law that states the limit of a product is equal to the product of the limits, and the fact that you should have been taught already that $\displaystyle \lim_{u \to 0} \dfrac{\sin u}{u} = 1$.

Thanks.