Limit x tends to 0 1-√cos x/ x^2 evaluate the limit
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Answered by
26
the answer is 1/4
refer to the image attached for solution
refer to the image attached for solution
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tokaians:
why did you divide and multiply x^2 with 4
lim(sinx/x) as x-->0 is 1 so in the numerator there we have sin^2(x/2); for each sin(x/2) we divide by x/2 and for sin^2(x/2) we divide by x^2/4;since we only have x^2 in the denominator so to bring (x^2/4) i have multiplied and divided by 4
if still problem persists then pls inform
:)
so when we have x/2 we divide an multiply with x/2 and when we have sin^2 x/2 we divide and multiply with x^2/4
we ultimately bring in the form sin(A)/A as A tends to 0
so we multiply and divide accordingly
we also have the L hospitals rule...but do not go for it.....use it only when all other methods fail
yes you r right ✌ it is decreases your thinking capability
i gave it coz you have answered in the other way
Answered by
8
the another method to solve o/o form is l-hospital rule
lim x→o 1-√cosx / x²
applying l hospital rule
limx→0 d/dx(1- √cosx) / d/dx(x²)
lim x→0 sinx/2√cosx / 2x
=> lim x→0 sinx/4x√cosx
we know that lim x→0 sinx/x = 1 and lim x→0 cos x = 1
apply it here we get
lim x→0 1/4
ask we if you will face any problem regarding this
lim x→o 1-√cosx / x²
applying l hospital rule
limx→0 d/dx(1- √cosx) / d/dx(x²)
lim x→0 sinx/2√cosx / 2x
=> lim x→0 sinx/4x√cosx
we know that lim x→0 sinx/x = 1 and lim x→0 cos x = 1
apply it here we get
lim x→0 1/4
ask we if you will face any problem regarding this
we differentiate numerator and denominator separately
until zero by zero form is disappeared
ok
ye understand ?
yes
so what is the differentiation of 1-√cosx
0 - 1/2(-sinx)/root cosx
chain rule
okkk
:) ask me if you will face any in this chapter
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