limit 0 tends to 90° sin^4x cos^3x dx
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answer is lim_(xrarr0)(cos(3x)-cos(4x))/x^2=7/2
Explanation:
If you try to answer this question by direct substitution, you get:
lim_(xrarr0)(cos(3x)-cos(4x))/x^2=
lim_(xrarr0)(cos(3*0)-cos(4*0))/0^2
=(1-1)/0=0/0
Since you cannot divide by zero, it is tempting to say that there is no solution. However, 0/0 (or oo/oo or a number of other seemingly impossible results obtained when finding limits) is actually called an indeterminate form. It does not automatically mean that there is no limit. It means there may be no limit, but we must investigate further. Sometimes we use algebra.
This time we will use L'Hhatopital's Rule, which says that for a number of cases with indeterminate forms, if:
lim_(xrarra)(f(x))/(g(x))=0/0 then
lim_(xrarra)(f(x))/(g(x))=lim_(xrarra)(f'(x))/(g'(x))
So, we take the derivative of the numerator and the derivative of the denominator. Note that we are not taking the derivative of the entire fraction so we do not use the quotient rule!
lim_(xrarr0)(3sin(3x)-4sin(4x))/(2x)
Note the use of the chain rule in the two terms of the numerator. This gives us
lim_(xrarr0)(3sin(3x)-4sin(4x))/(2x)=
lim_(xrarr0)(3sin(3*0)-4sin(4x*0))/(2*0)=
Explanation:
If you try to answer this question by direct substitution, you get:
lim_(xrarr0)(cos(3x)-cos(4x))/x^2=
lim_(xrarr0)(cos(3*0)-cos(4*0))/0^2
=(1-1)/0=0/0
Since you cannot divide by zero, it is tempting to say that there is no solution. However, 0/0 (or oo/oo or a number of other seemingly impossible results obtained when finding limits) is actually called an indeterminate form. It does not automatically mean that there is no limit. It means there may be no limit, but we must investigate further. Sometimes we use algebra.
This time we will use L'Hhatopital's Rule, which says that for a number of cases with indeterminate forms, if:
lim_(xrarra)(f(x))/(g(x))=0/0 then
lim_(xrarra)(f(x))/(g(x))=lim_(xrarra)(f'(x))/(g'(x))
So, we take the derivative of the numerator and the derivative of the denominator. Note that we are not taking the derivative of the entire fraction so we do not use the quotient rule!
lim_(xrarr0)(3sin(3x)-4sin(4x))/(2x)
Note the use of the chain rule in the two terms of the numerator. This gives us
lim_(xrarr0)(3sin(3x)-4sin(4x))/(2x)=
lim_(xrarr0)(3sin(3*0)-4sin(4x*0))/(2*0)=
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