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in Limits by (30 points)

If L = \(\lim\limits_{x \to 0} \frac{asinx-bx+cx^2+x^3}{2x^2log(1+x)-2x^3+x^4}\)exists and it is finite then find a, b, c and L

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by (55.0k points)

\(\lim\limits_{x \to 0}\,\cfrac{a\,sinx-bx+cx^2+x^3}{2x^2log(1+x)-2x^3+x^4}\)

\(=\lim\limits_{x \to 0} \,\cfrac{a(x-\frac{x^3}{3!}+\frac{x^5}{5!}....)-bx+cx^2+x^3}{2x^2(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^2}{4})-2x^3+x^4}\)

\(=\lim\limits_{x \to 0} \, \cfrac{(a-b)x+cx^2-(\frac{a}{b}-1)x^3+\frac{a}{120}x^5...}{2x^3-x^4+2x^2 (\frac{x^3}{3}-\frac{x^4}{4}+....)-2x^3+x^ 4}\)

\(=\lim\limits_{x \to 0} \,\cfrac{(a-b)x+cx^2-(\frac{a}{b}-1)x^3+\frac{a}{120}x^ 5}{2x^5(\frac{1}{3}-\frac{x}{4}+...)}\)

It will be finite if,

a-b = 0, c = 0, a/b - 1 = 0

ie = a = b, c = 0, a = 6

⇒ a = b = 6, c = 0

\(L=\frac{\frac{a}{120}}{\frac{2}{3}}=\frac{a}{120}\times \frac{3}{2} =\frac{a}{80} =\frac{6}{80}=\frac{3}{40}.\)

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