Evaluate the limit of the function ln(1+x)/(sinx+sin3x) x-->0
- print Print
- list Cite
Expert Answers
calendarEducator since 2010
write12,544 answers
starTop subjects are Math, Science, and Business
We have to find the value of lim x--> 0 [ ln(1+x)/(sinx+sin3x)]
substituting x = 0, we get the indeterminate form 0/0. Therefore we can use l'Hopital's Rule and substitute the numerator and denominator with their derivatives.
=> lim x--> 0 [ (1/(1+x))/(cos x + 3*cos 3x)]
Substitute x = 0
[ (1/(1+x))/(cos x + 3*cos 3x)]
=> (1 / 1) / ( 1 + 3)
=> 1/4
The required value of lim x--> 0 [ln(1+x)/(sinx+sin3x)] = (1/4)
Related Questions
- Determine the limit of the function (sin5x-sin3x)/x, x-->0
- 1 Educator Answer
- Verify if limit of ln(1+x)/x is 1, x-->0
- 1 Educator Answer
- Evaluate the limit of the function (2x-sin2x)/x^3; x-->0.
- 1 Educator Answer
- Prove that limit of the function (a^x-1)/x=lna,x->0,using two methods.
- 1 Educator Answer
- How to evaluate the limit of (cos x - cos 3x) / x*sin x if x-->0 ?
- 1 Educator Answer
First, we'll substitute x by 0 into the function:
lim ln(1+x)/(sinx+sin3x) = ln 1/(sin 0 + sin 0) = 0/0
We've get an indetermination.
Since the trigonometric functions from denominator are matching, we'll transform the sum into a product:
sinx+sin3x = 2[sin (x+3x)/2][cos(x-3x)/2]
sinx+sin3x = 2sin 2x*cos(-x)
Since the cosine function is even, we'll put cos(-x) = cos x
sinx+sin3x = 2sin 2x*cos x
We'll re-write the limit:
lim ln(1+x)/(sinx+sin3x) = lim ln(1+x)/2sin 2x*cos x
We'll re-arrange the terms, crating remarcable limits:
(1/2)*lim ln(1+x)/sin 2x*lim (1/cos x)
But lim (1/cos x) = 1/cos 0 = 1
(1/2)*lim ln(1+x)/sin 2x= (1/2)lim [ln(1+x)/x]*(2x/sin 2x)*(x/2x)
(1/2)lim [ln(1+x)/x]*(2x/sin 2x)*(x/2x) = (1/2)*1*1*lim (x/2x)
(1/2)lim (x/2x) = 1/4
For x->0, lim ln(1+x)/(sinx+sin3x) = 1/4.
Unlock This Answer Now
Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.
Student Answers