LimitsHard

Question

If $\lim_{x \rightarrow 0}\mspace{2mu}\frac{e^{(a - 1)x} + 2cosbx + (c - 2)e^{- x}}{xcosx - \log_{e}(1 + x)} = 2$, then $a^{2} + b^{2} + c^{2}$ is equal to :

Options

A.5
B.3
C.7
D.9

Solution

$\lim_{x \rightarrow 0}\mspace{2mu}\frac{\left( 1 + (a - 1)x + \frac{(a - 1)^{2}x^{2}}{2!} \right) + 2\left( 1 - \frac{b^{2}x^{2}}{2!} \right) + (c - 2)\left( 1 - x + \frac{x^{2}}{2!} \right)}{x\left( 1 - \frac{x^{2}}{2!} \right) - \left( x - \frac{x^{2}}{2}\ldots \right)} = 2$

$$\lim_{x \rightarrow 0}\mspace{2mu}\frac{(1 + 2 + c - 2) + x(a - 1 - c + 2) + x^{2}\left( \frac{(a - 1)^{2}}{2} - b^{2} + \left( \frac{c - 2}{2} \right) \right)}{\frac{x^{2}}{2} - \frac{x^{3}}{2!} + \ldots} = 2 $$For which

$${\because c + 1 = 0 \Rightarrow c = - 1 }{\because a - c = - 1 \Rightarrow a = - 2 }{\because\frac{(a - 1)^{2}}{2} - b^{2} + \left( \frac{c - 2}{2} \right) = 1 }{\frac{9}{2} - b^{2} - \frac{3}{2} = 1 \Rightarrow {\text{ }b}^{2} = 2 }{a^{2} + b^{2} + c^{2} = 4 + 2 + 1 = 7}$$

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