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

We have to evaluate
$$L=\lim_{x \rightarrow 0}\dfrac{e^{(a-1)x}+2\cos bx+(c-2)e^{-x}}{x\cos x-\ln(1+x)}$$
and the question states that $$L=2$$. We shall expand the numerator and the denominator in Maclaurin series up to $$x^{2}$$ terms.

Series expansion of the numerator

1. $$e^{(a-1)x}=1+(a-1)x+\dfrac{(a-1)^{2}}{2}\,x^{2}+O(x^{3})$$
2. $$2\cos bx=2\Bigl(1-\dfrac{b^{2}x^{2}}{2}+O(x^{4})\Bigr)=2-b^{2}x^{2}+O(x^{4})$$
3. $$(c-2)e^{-x}=(c-2)\Bigl(1-x+\dfrac{x^{2}}{2}+O(x^{3})\Bigr)=(c-2)-(c-2)x+\dfrac{(c-2)}{2}x^{2}+O(x^{3})$$

Adding the three expressions:

Constant term: $$1+2+(c-2)=c+1$$
Coefficient of $$x$$: $$(a-1)-(c-2)=a-c+1$$
Coefficient of $$x^{2}$$: $$\dfrac{(a-1)^{2}}{2}-b^{2}+\dfrac{(c-2)}{2}=\Bigl[\dfrac{(a-1)^{2}}{2}-b^{2}+\dfrac{c-2}{2}\Bigr]$$

Hence $$e^{(a-1)x}+2\cos bx+(c-2)e^{-x}=(c+1)+(a-c+1)x+\Bigl[\dfrac{(a-1)^{2}}{2}-b^{2}+\dfrac{c-2}{2}\Bigr]x^{2}+O(x^{3}).$$

Series expansion of the denominator

$$\cos x=1-\dfrac{x^{2}}{2}+O(x^{4})\;\;\Longrightarrow\;\;x\cos x=x-\dfrac{x^{3}}{2}+O(x^{5})$$
$$\ln(1+x)=x-\dfrac{x^{2}}{2}+\dfrac{x^{3}}{3}+O(x^{4})$$
Therefore

$$x\cos x-\ln(1+x)=\Bigl[x-\dfrac{x^{3}}{2}\Bigr]-\Bigl[x-\dfrac{x^{2}}{2}+\dfrac{x^{3}}{3}\Bigr]+O(x^{4}) =\dfrac{x^{2}}{2}-\dfrac{x^{3}}{6}+O(x^{4}).$$

The denominator starts with $$\dfrac{x^{2}}{2}$$, so for the limit to be finite the numerator must also start with an $$x^{2}$$ term. Hence

Constant term condition: $$c+1=0\;\;\Longrightarrow\;\;c=-1$$
Linear term condition: $$a-c+1=0\;\;\Longrightarrow\;\;a-(-1)+1=0\;\;\Longrightarrow\;\;a=-2$$

With $$a=-2,\;c=-1$$, the $$x^{2}$$ coefficient of the numerator becomes

$$\dfrac{(a-1)^{2}}{2}-b^{2}+\dfrac{c-2}{2} =\dfrac{(-3)^{2}}{2}-b^{2}+\dfrac{-3}{2} =\dfrac{9}{2}-b^{2}-\dfrac{3}{2}=3-b^{2}.$$

The denominator’s $$x^{2}$$ coefficient is $$\dfrac{1}{2}$$. Thus the required limit equals

$$L=\dfrac{3-b^{2}}{1/2}=2(3-b^{2}).$$

Given that $$L=2$$, we get
$$2(3-b^{2})=2\;\;\Longrightarrow\;\;3-b^{2}=1\;\;\Longrightarrow\;\;b^{2}=2.$$

Value of $$a^{2}+b^{2}+c^{2}$$

$$a=-2\;\Rightarrow\;a^{2}=4,$$ $$b^{2}=2,$$ $$c=-1\;\Rightarrow\;c^{2}=1.$$
Therefore $$a^{2}+b^{2}+c^{2}=4+2+1=7.$$

Hence the required value is 7 (Option B).