Question #d5a95

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#