Ask a New Question

Question

Find the limit of x as it approaches zero. (Sin^2 8x)/x^2
12 years ago

Answers

Sanji
lim ((sin 8x)^2)/x^2 as x->0
since as x->0 the denominator becomes zero, use l'hopital's rule:
lim (16 sin(8x)cos(8x) / 2x)
lim (8 sin(16x))/2x
another l'hopital's rule:
lim (128 cos (16x))/2
as x->0,
(128 * cos(0))/2
= 64
12 years ago
Damon
as x --> 0 sin (8x)^2---> 64 x^2
64 x^2/x^2 = 64
12 years ago

Related Questions

Find the limit Limit as h approaches 0 of : SqRt(4+h)-2 ____________ h by rela... (a) Find the tangent line approximation for sqrt(9+x) near x=0. (b)Find a formula for the error,... Prove that the limit as n approaches infinity of ((n!)^2)/(2n)! is 0. I'm thinking that you have... Find the limit limit as x approaches 1 of (2-x)^3tan[(pi/2)x] proof that the limit as x approaches 2 of x to the 4th equals 16 Show that the limit as h approaches 0 of (e^h-1 ) all over h = ln e = 1 using at least two numerica... Given that the limit as h approaches 0 of (f(6 + h) - f(6))/h = -2, which of these statements must b... Find the limit. (If the limit is infinite, enter '∞' or '-∞', as appropriate. If the limit does not... We’re trying to find an approximation for 20 ​ . Now that you’ve found the perfect squares, l... We’re trying to find an approximation for 20 ​ . Between which two perfect squares is 20? Pe...
Ask a New Question
Archives Contact Us Privacy Policy Terms of Use