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Answer 1 · 2018-02-07T13:10:10

Answer 2 · 2018-02-07T14:48:40

See the 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:

$lim_(x to 0) [sin(x)/x]$

When we evaluate this limit we get $sin(0)/0 = 0/0$ , an indeterminate form.

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

$lim_(x to 0) [[sin(x)']/[(x)']]$

Differentiate $sin(x)$

$(dy)/(dx) sin(x) = cos(x)$

and

$(dy)/(dx) (x) = 1$

We will use these intermediate results in

$lim_(x to 0) [[sin(x)']/[(x)']]$

We get,

$lim_(x to 0) [[cos(x)]/[1]]$

$rArr lim_(x to 0) cos(x)$

Substitute $x = 0$ to get

$Cos(0) = 1$

Hence,

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