Question d5a95

Feb 7, 2018

See the explanation.

Explanation:

We must prove that

color(brown)(lim_(x to 0)sin(x)/x =1

L'Hospital's Rule is a handy tool for calculating limits involving indeterminate forms like

color(green)(0/0 or (+- oo)/(+-oo)

If we are taking the limit

color(blue)(lim_(x to a)f(x)/g(x), we get color(blue)(f(a)/g(a) = 0/0 or (+- oo)/(+-oo)

Apparently, when we use substitution method to evaluate the limit, we end up with an indeterminate form

L'Hospital's Rule states that instead of evaluating

color(blue)(lim_(x to a)f(x)/g(x)

we can evaluate the limit of

color(blue)(lim_(x to a)(f'(x))/(g'(x)

Given:

${\lim}_{x \to 0} \left[\sin \frac{x}{x}\right]$

When we evaluate this limit we get $\sin \frac{0}{0} = \frac{0}{0}$, an indeterminate form.

Hence, we will use the L'Hospital's Rule for our problem.

${\lim}_{x \to 0} \left[\frac{\sin \left(x\right) '}{\left(x\right) '}\right]$

Differentiate $\sin \left(x\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} \sin \left(x\right) = \cos \left(x\right)$

and

$\frac{\mathrm{dy}}{\mathrm{dx}} \left(x\right) = 1$

We will use these intermediate results in

${\lim}_{x \to 0} \left[\frac{\sin \left(x\right) '}{\left(x\right) '}\right]$

We get,

${\lim}_{x \to 0} \left[\frac{\cos \left(x\right)}{1}\right]$

$\Rightarrow {\lim}_{x \to 0} \cos \left(x\right)$

Substitute $x = 0$ to get

$C o s \left(0\right) = 1$

Hence,

color(brown)(lim_(x to 0)sin(x)/x =1#