What is the limit? lim_(xrarr2) (cos(pi/x))/(x-2)
3 Answers
Explanation:
Since we get
L'Hospital's Rule:
I am assuming you know the trigonometric relationships, the power rule, and the chain rule.
We now plug 2 in the place of
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
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
Without L'Hospital's rule see below.
Explanation:
= sin((pi(x-2))/(2x))/(x-2)
= pi/(2x) (sin((pi(x-2))/(2x)))/((pi(x-2))/(2x)
As