What is the limit? #lim_(xrarr2) (cos(pi/x))/(x-2)#

Since we get #0/0# after plugging #2# in the place of #x#, we can us the L'Hospital's Rule.

L'Hospital's Rule: #lim_(x->c)(f(x))/(g(x))=(f'(c))/(g'(c))# when you get #f(c)# gives you an indeterminate form.

#=>(d/dx(cos(pi/x)))/(d/dx(x-2))#

I am assuming you know the trigonometric relationships, the power rule, and the chain rule.

#=>(-sin(pi/x)*d/dx(pix^(-1)))/1#

#=>-sin(pi/x)*-pix^(-2)#

#=>sin(pi/x)pix^-2#

We now plug 2 in the place of #x#.

#=>sin(pi/2)pi*2^-2#

#=>1*pi*1/4#

#=>pi/4#

We seek:

# L = lim_(xrarr2) (cos(pi/x))/(x-2) #

Method 1:

Both the numerator or denominator are zero at #x=2#, and so we have an indeterminate form #0/0#and we can apply L'Hôpital's rule, which states that if we have an indeterminate form, and the limit exists, then:

# lim_(x rarr a) f(x)/g(x) = lim_(x rarr a) (f'(x))/(g'(x)) #

And so:

# L = lim_(xrarr2) (d/dx cos(pi/x))/(d/dx x-2) #
# \ \ \ = lim_(xrarr2) (-sin(pi/x) * ( -pi/x^2) )/(1) #
# \ \ \ = lim_(xrarr2) (pi sin(pi/x) /x^2) #
# \ \ \ = pi sin(pi/2) /2^2 #
# \ \ \ = (pi) /4 #

#cos A = sin (pi/2-A)# so we have

#cos(pi/x)/(x-2) = sin(pi/2-pi/x)/(x-2)#

# = sin((pi(x-2))/(2x))/(x-2)#

# = pi/(2x) (sin((pi(x-2))/(2x)))/((pi(x-2))/(2x)#

As #xrarr2#, we have #((pi(x-2))/(2x))rarr0# so we get

#lim_(xrarr2)pi/(2x) (sin((pi(x-2))/(2x)))/((pi(x-2))/(2x)) = pi/(2(2)) (1) = pi/4#