Calculus
posted by Seeina
Determine the limit
lim x>0 1cos^2(3x)/x^2

MathMate
lim x>0 1cos^2(3x)/x^2
Since it is a situation of 0/0, we can use l'Hôpital's rule by differentiating both numerator and denominator to get:
lim x>0 1cos^2(3x)/x^2
=lim x>0 6sin(3x)/2x
Since it is still a situation of 0/0, we can apply l'Hôpital's rule again:
=lim x>0 18cos(3x)/2
= 18*1/2
= 9
If you have learned about Taylor series or MacLauin series, you can expand the numerator into a series, and evaluate accordingly:
lim x>0 1cos^2(3x)/x^2
=lim x>0 1(1(3x)^2/2!+(3x)^4/4!...)^2/x^2
=lim x>0 1(19x^2+27x^4...)/x^2
=lim x>0 (9x^227x^4)/x^2
=lim x>0 (927x^2+...)
=9
Respond to this Question
Similar Questions

Calculus
yes! tnk u ok? It's actually (x>0.) Find the limit of cot(x)csc(x) as x approached 0? 
CALCULUS  need help!
Determine the limit of the trigonometric function (if it exists). 1. lim sin x / 5x (x > 0) 2. lim tan^2x / x (x >0) 3. lim cos x tan x / x (x > 0) 
calculus
Determine the infinite limit of the following function. Lim as x>zero 1/x^2(x+7) and lim as x>3 from the left side 2/x3 
calculus
The limit represents the derivative of some function f at some number a. Select an appropriate f(x) and a. lim (cos(pi+h)+1)/h h>0 answers are f(x) = tan(x), a = pi f(x) = cos(x), a = pi/4 f(x) = cos(x), a = pi f(x) = sin(x), a … 
calculus
calculate the lim as x> pi (cos(x)+1)/(xpi) using the special limit lim x>0 sinx/x 
Calculus
evaluate the limit: lim cos(x + pi/2)/x x>0 
Calculus. Limits. Check my answers, please! :)
4. lim (tanx)= x>pi/3 (sqrt3) 1 (sqrt3) ***1 The limit does not exist. 5. lim x= x>2 2 ***2 0 1 The limit does not exist. 6. lim [[x]]= x>9/2 (Remember that [[x]] represents the greatest integer function of x.) 4 … 
Calculus help, please!
1. Evaluate: lim x>infinity(x^47x+9)/(4+5x+x^3) 0 1/4 1 4 The limit does not exist. 2. Evaluate: lim x>infinity (2^x+x^3)/(x^2+3^x) 0 1 3/2 2/3 The limit does not exist. 3. lim x>0 (x^37x+9)/(4^x+x^3) 0 1/4 1 9 The limit … 
Calculus, please check my answers!
1. Evaluate: lim x>infinity(x^47x+9)/(4+5x+x^3) 0 1/4 1 4 ***The limit does not exist. 2. Evaluate: lim x>infinity (2^x+x^3)/(x^2+3^x) 0 1 3/2 ***2/3 The limit does not exist. 3. lim x>0 (x^37x+9)/(4^x+x^3) 0 1/4 1 ***9 … 
Limits
Let lim f(x) =16 as x>4 and lim g(x) =8 as x>4 Use the limit rules to find lim [cos(pi*f(x)/g(x))] as x>4