The limit as $x$ goes to infinity of $x-x\cos(4/x).$

Question

I would like to determine $$\lim_{x\to \infty} \left(x-x\cos \frac4x\right)$$

How do I even start this? I can't plug in infinity to $4/x,$ can I? That will make it zero? Is the answer infinity then?

Answer 1

Notice, you can put $ t = \frac{1}{x} $. Hence

$$ \lim_{x \to \infty} x - x \cos(\frac{4}{x} ) = \lim_{t \to 0} \frac{1}{t} - \frac{1}{t} \cos(4t) = \lim_{t \to 0} \frac{ 1 - \cos (4t) }{t} $$

Now, multiply this by $\frac{4}{4} $ and remember the following well known limit:

$$ \lim_{\alpha \to 0 } \frac{ 1 - \cos \alpha}{\alpha} = 0 $$

Answer 2

Notice that if $u\to 0$ then

$$\cos u=1+\mathcal O(u^2)$$ so we get

$$x-x\cos\left(\frac4x\right)=x-x+\mathcal O\left(\frac1x\right)\xrightarrow{x\to\infty}0$$

Answer 3

Putting $x=\frac{1}{t}$ it becomes:

$$\ \lim_{t\to0}\frac{1-\cos(4t)}{t}=\lim_{t\to0}\frac{1-\cos(4t)}{16t^2}\cdot\frac{16t^2}{t}=\lim_{t\to0}8\cdot t=0$$

Answer 4

We can write $$x-x\cos\frac4x=x\frac{1-\cos^2\frac4x}{1+\cos\frac4x}=\left(\frac{\sin\frac4x}{\frac4x}\right)^2\cdot \frac{16}{1+\cos\frac4x}\cdot\frac1x.$$ Now as $x\to\infty$, the limit of the first factor is $1$, the limit of the second is $8$, and the limit of the third is $0$. I let you conclude.

Answer 5

Hint: Factor x, and use the fact that $~\dfrac{1-\cos t}2=\sin^2\dfrac t2~$ in conjunction with $~\displaystyle\lim_{u\to0}\dfrac{\sin u}u=1$.