If a,b,c,d are constants such that $\displaystyle \displaystyle\lim_{x\to{0}}\frac{ax^2+sin\;bx+sin\ ;cx+sin\;dx}{2x^2+3x^3+4x^4} = 3$ what is the value of the sum a+b+c+d? Thanks!
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Use L'Hopital's rule. you get (0+b+c+d)/0 I think the only way this limit can exist is if b+c+d = 0 we then get the indeterminate form 0/0 Use L'Hopital's rule again. we get 2a/4 as the limit so a/2 = 3 and a = 6.
Originally Posted by jenius Use L'Hopital's rule. you get (0+b+c+d)/0 I think the only way this limit can exist is if b+c+d = 0 we then get the indeterminate form 0/0 Use L'Hopital's rule again. we get 2a/4 as the limit so a/2 = 3 and a = 6. So is 6 the final answer?
Unless I made a mistake somewhere, 6 is the final answer.
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