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  • Given   <img alt="\mathrm{\lim _{x \rightarrow \infty} x\left(2+(3+x) \ln \left(\frac{x+a}{x+b}\right)\right)=2}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Clim%2

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 }}


Answers (1)

best_answer

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.}

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