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

3 Answers
Feb 19, 2018

lim_(x->2)cos(pi/x)/(x-2)=pi/4

Explanation:

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

Feb 19, 2018

lim_(xrarr2) (cos(pi/x))/(x-2) = (pi) /4

Explanation:

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 0/0and 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

Feb 19, 2018

Without L'Hospital's rule see below.

Explanation:

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