Question

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

Answer 1

`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`

Answer 2

`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 `0/0`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`

Answer 3

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`