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  • If  <img alt="\mathrm{\lim _{x \rightarrow 0}\left[\frac{a x e^x-b \log (1+x)+c x e^{-x}}{x^2 \sin x}\right]=2}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Clim%20_%7Bx

If  \mathrm{\lim _{x \rightarrow 0}\left[\frac{a x e^x-b \log (1+x)+c x e^{-x}}{x^2 \sin x}\right]=2}, then the value of \mathrm{(a, b, c)} is

Option: 1

(12,9,3)


Option: 2

(3,9,12)


Option: 3

(9,3,12)


Option: 4

(3,12,9)


Answers (1)

best_answer

Using expansion, we have

\mathrm{\lim _{x \rightarrow 0} a x\left(1+x+\frac{x^2}{2 !}+\ldots\right)-b\left(x-\frac{x^2}{2}+\frac{x^2}{3}-\ldots\right)}

\mathrm{\frac{+c x\left(1-\frac{x}{1 !}+\frac{x^2}{2 !}-\frac{x^3}{3 !}-\cdots\right)}{x^2\left(x-\frac{x^3}{3 !}+\frac{x^5}{5 !}-\ldots\right)}=2}

\mathrm{\begin{gathered} \Rightarrow \frac{\lim _{x \rightarrow 0} x(a-b+c)+x^2\left(a+\frac{b}{2}-c\right)+x^3\left(\frac{a}{2}-\frac{b}{3}+\frac{c}{2}\right)+\ldots}{x^2-\left(x-\frac{x^3}{3 !}+\frac{x^5}{5 !}-\ldots\right)} \\ =2 \end{gathered}}

Now, above limit would exist, if least power in numerator is greater than or equal to least power in denominator, i.e. coefficient of x or \mathrm{x^2} must be zero and coefficient of \mathrm{x^3} should be 2.

\mathrm{\therefore a-b+c=0, a+\frac{b}{2}-c=0 \text { and } \frac{a}{2}-\frac{b}{3}+\frac{c}{2}=2}

On solving, we get

                      \mathrm{a=3, b=12 \text { and } c=9}

\mathrm{\therefore(a, b, c)=(3,12,9)}

Posted by

manish painkra

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