#### Given   $\mathrm{\lim _{x \rightarrow \infty} x\left(2+(3+x) \ln \left(\frac{x+a}{x+b}\right)\right)=2}$Option: 1 $\mathrm{a=3, b=5}$Option: 2 $\mathrm{a=2, b=4}$Option: 3 $\mathrm{a=10, b=7}$Option: 4 $\mathrm{\text { None of these }}$

The given limit condition implies that $\mathrm{(x+3) \log ((x+a) /(x+b)) \rightarrow-2}$ Then clearly $\mathrm{a \neq b}$ and we have

$\mathrm{ (a-b) \cdot \frac{x+3}{x+b} \cdot \frac{\log \left(1+\frac{a-b}{x+b}\right)}{\frac{a-b}{x+b}} \rightarrow-2 }$
Thus $\mathrm{a-b=-2}$ or $\mathrm{a=b-2.}$ Using this value of a in original limit we see that

$\mathrm{ x\left(2+(x+3) \log \left(1-\frac{2}{x+b}\right)\right) \rightarrow 2 }$

Replacing the first factor by $x+b$ and putting $x+b=2 / t$ we see that

$\mathrm{ \lim _{t \rightarrow 0^{+}} \frac{1}{t}\left(2+\left(\frac{2}{t}+3-b\right) \log (1-t)\right)=1 }$
or

$\mathrm{\frac{2 t+2 \log (1-t)}{t^2}+(3-b) \frac{\log (1-t)}{t} \rightarrow 1}$

By L'Hospital's Rule or Taylor series the first fraction tends to -1 so that $\mathrm{-1+b-3=1}$ or $\mathrm{b=5}$ and $\mathrm{a=3.}$