# How do you find the limit of (x-cosx)/x as x approaches oo?

May 22, 2017

$1$

#### Explanation:

${\lim}_{x \to \infty} \left[\frac{x - \cos x}{x}\right] = {\lim}_{x \to \infty} \left[\frac{x}{x} - \cos \frac{x}{x}\right]$
$= {\lim}_{x \to \infty} \left[1 - \cos \frac{x}{x}\right] = {\lim}_{x \to \infty} \left[1\right] - {\lim}_{x \to \infty} \left[\cos \frac{x}{x}\right]$

$\to {\lim}_{x \to \infty} \left[\cos \frac{x}{x}\right] = 0$

The range of $\cos x$ is $- 1 \le \cos x \le 1$, however as $x \to \infty$ you are dividing a small number by an increasingly large number .

Thus:
$= 1 - 0 = 1$

May 22, 2017

lim_(x->∞ )(x-cosx)/x=1

#### Explanation:

Firstly, the limit of a sum is the sum of the limits

lim_(x->∞ )(f(x)+g(x))=lim_(x->∞ )f(x)+lim_(x->∞ )g(x)

Separate the terms and solve them individually

lim_(x->∞ )(x-cosx)/x=lim_(x->∞ )x/x-lim_(x->∞)cosx/x

The first term is

lim_(x->∞ )x/x=∞/∞

Use L'Hôpital's Rule for such limits of indeterminate form

lim_(x->∞ )f(x)/g(x)=lim_(x->∞ )f^'(x)/g^'(x)=L

In this case

lim_(x->∞ )x/x=lim_(x->∞ )1/1=1

Use the Squeeze or Sandwich Theorem for the second term. This is more complicated and involves three steps. We will apply them in order.

If

$g \left(x\right) \le f \left(x\right) \le h \left(x\right)$ (1)

and

lim_(x->∞)g(x)=lim_(x->∞)h(x)=L (2)

then

lim_(x->∞)f(x)=L (3)

(1) Let
$f \left(x\right) = \frac{1}{x} \cdot \cos x$
$g \left(x\right) = - \frac{1}{x}$
$h \left(x\right) = \frac{1}{x}$

We know that cos(x) goes from -1 to 1

$- 1 \le \cos x \le 1$

For f(x), we are multiplying 1/x by cos(x), which means multiplying by numbers from -1 to 1.
The upper and lower bounds of f(x) can now be found. The maximum value the function can take is 11/x and the minimum will be -11/x. Multiplying 1/x by any other number in-between -1 and 1 will result in a smaller number within these bounds. This means that the following is true:

$- \frac{1}{x} \le \frac{1}{x} \cdot \cos x \le \frac{1}{x}$

This is saying that $\frac{1}{x} \cdot \cos x$ is squeezed or sandwiched between $- \frac{1}{x}$ and $\frac{1}{x}$ as x approaches infinite. You can see this when f(x), g(x) and h(x) are plotted together.

From the graph, you can see that as x approaches ∞, f(x) is being squeezed toward 0 by $\frac{1}{x}$ and $- \frac{1}{x}$. You can also see that $g \left(x\right) \le f \left(x\right) \le h \left(x\right)$ , that is, f(x) is contained within the bounds of g(x) and h(x).

(2) Although I just gave the answer, pretend that you didn't have that graph and move on to step two. We need to find the limits of g(x) and h(x) and see if they are equal.

lim_(x->∞)(1/x)=lim_(x->∞)(-1/x)=0=L

(3) They are equal, so from the Squeeze Theorem

lim_(x->∞)cosx/x=L=0

Finally, putting it all together

lim_(x->∞ )(x-cosx)/x=lim_(x->∞ )x/x-lim_(x->∞)cosx/x=1-0=1

You prove this for yourself by looking at the graph