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lim x-->0 [cos(sinx)-cosx] / x4

ans: 1/6

jyoti dogra , 13 Years ago
Grade 11
anser 4 Answers
Swapnil Saxena

Last Activity: 13 Years ago

Using L Hopitals rule

lim x tends to 0 [cos(sin(x))-cos(x)]/x^4 = d/dx[cos(sin(x))-cos(x)]/x^4 if ([cos(sin(x))-cos(x)]=0 and x=4)

Again since the expression is yielding 0/0 appyling L hopitals rule, 

lim x tends to 0 [cos(sin(x))-cos(x)]/x^4 = d2/dx2[cos(sin(x))-cos(x)]/x^4 at x=0

Again since the expression is yielding 0/0 appyling L hopitals rule, 

lim x tends to 0 [cos(sin(x))-cos(x)]/x^4 = d3/dx3[cos(sin(x))-cos(x)]/x^4 at x=0

Again since the expression is yielding 0/0 appyling L hopitals rule, 

lim x tends to 0 [cos(sin(x))-cos(x)]/x^4 = d4/dx4[cos(sin(x))-cos(x)]/x^4 at x=0

Now Since the expression 4/24= 1/6 .

Hence 1/6 is the answer 

 


Chetan Mandayam Nayakar

Last Activity: 13 Years ago

lim[cos(sinx)-cosx]/x^4

=lim[-sin(sinx)cosx+sinx]/(4x^3)

=lim[sin(sinx)sinx-cos(sinx)cos2x+cosx]/12x2

=lim[cosx-cos(sinx)cos2x]/12x2

=lim[1-cos(sinx)cosx]/12x2

=lim[sin(sinx)cos2x+cos(sinx)sinx]/24x

differentiate once more w.r.t. to get the answer

Dear Moderator: If I need to provide any more details, please contact me at <mn_chetan@yahoo.com

jyoti dogra

Last Activity: 13 Years ago

thanks a lot

Samyak Jain

Last Activity: 6 Years ago

Another method :
Using cosA – cosB = 2 sin(A + B)/2 sin(B – A)/2 
cos(sinx) – cosx = 2 sin{(sinx + x)/2} sin{(x – sinx)/2}
 
Let L be the given limit.
\therefore L = lim x\rightarrow0 [cos(sinx) – cosx] / x4 
       = lim x\rightarrow0 [2 sin{(sinx + x)/2} sin{(x – sinx)/2}] / x4
       = 2 . lim x\rightarrow0 sin{(sinx + x)/2} . {(sinx + x)/2} / (sinx + x)/2 . 
                    sin{(x – sinx)/2} . {(x – sinx)/2} / {(x – sinx)/2} / x4
 
As x\rightarrow0, (sinx + x)/2 \rightarrow0 and (x – sinx)/2 \rightarrow0.
L = 2 . lim x\rightarrow0 sin{(sinx + x)/2} / (sinx + x)/2 .lim x\rightarrow0 sin{(x – sinx)/2} / {(x – sinx)/2} . 
        lim x\rightarrow0 {(sinx + x)/2} . {(x – sinx)/2} / x4
 
We know that lim x\rightarrow0  sinx / x = 1
\therefore lim x\rightarrow0 sin{(sinx + x)/2} / (sinx + x)/2 = 1
and lim x\rightarrow0 sin{(x – sinx)/2} / {(x – sinx)/2} = 1
 
So, L = 2 . 1 . 1 . lim x\rightarrow0 {(sinx + x)/2} . {(x – sinx)/2} / x4
          = 2 lim x\rightarrow0 (sinx + x) . (x – sinx) / 4x4
          = ½ im x\rightarrow0 (x2 – sin2x) / x4
Here use expansion of sine series.
sinx = x – x3 / 3! + x5 / 5! – ….
sin2x = x2 – 2x4 / 3! + f(x), where degree of x in f(x) is greater than 4.
\therefore L = ½ lim x\rightarrow0 [ x2 – {x2 – 2x4 / 3! + f(x)} ] / x4
L = ½ lim x\rightarrow0 [ x2 – x2 + 2x4 / 6 –  f(x)} ] / x4
L = ½ lim x\rightarrow0 [ x4 / 3 –  f(x)} ] / x4
L = ½ lim x\rightarrow0 [ x4 / 3x4 ] –  lim x\rightarrow0 [ f(x) / x4 ]
\because f(x) contains powers of x greater than 4,  lim x\rightarrow0 [ f(x)} ] / x4 = 0,
 L = ½ lim x\rightarrow0 [ x4 / 3x4 ] = ½ . 1/3 = 1/6
\therefore lim x\rightarrow0 [cos(sinx) – cosx] / x4 = 1/6

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