can anyone solve this difficult limit problem?????

Question

lim((sinпx-cosпx)/(16 * x ^ 2 -1)) when x --->0.25
(п=pi)

Answer 1

LH rule
limit = π*(cosπx + sinπx) / (32x)
= π*(2/√2) / 8
= π√2 / 8

Answer 2

(pi*sqrt(2))/8, or 0.55536

If you evaluate it as is, you get 0/0. Thus, you need to use l'Hôpital's rule and take the derivatives of the numerator and denominator; these become the new numerator and denominator. The new fraction is (ncosnx+nsinnx)/(32x) where n=pi. Then take the limit of the new fraction as x --->0.25, (this limit is equal to the limit of the original), and the answer you get is (pi*sqrt(2))/8, or 0.55536

Answer 3

We can find this limit with the Hospital' rule :

the limit of the numerator and the denominator are 0, then the limit is equal to the limit of :

(sin(pi x) - cos(pi x))' / (16x²-1)'
= pi(cos(pi x) + sin(pi x)) / 32x

so, the limit is pi[sqrt(2)/2+ sqrt(2)/2] / (32/4)
=pi sqrt(2)/8

Answer 4

For difficult limits try l'hopital's rule i.e. look at f'(x) / g'(x)

Here f'(x) = pi cos(pix) + pi sin(pix) and
g'(x) = 32x
now limit of f'(x) / g'(x) = pi*sqrt(2) / 8

Answer 5

Nope, answer is pi/(4 sqrt(2))

Use l'Hopital's Rule

d/dx (sin (pi x) - cos(pi x)) = pi(cos(pi x) -sin(pi x))

d/dx(16x^2 -1) = 32x

Substitute x = 1/4

= pi/sqrt(32)

Answer 6

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Answer 7

If you know of L"Hospital's Rule then it is the same limit as [n cos(nx) + n sin(nx)] / [ 32x]....n sqrt2 / 8

Answer 8

yes thats so easy its your mom

Answer 9

it is solved by direct substitution.the answer is zero