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  • If <img alt="\mathrm{\operatorname{Lt}_{x \rightarrow \infty}\left[\frac{x^2+1}{x+1}-a x-b\right]=\infty}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Coperatorname%7BLt%7D_%

If \mathrm{\operatorname{Lt}_{x \rightarrow \infty}\left[\frac{x^2+1}{x+1}-a x-b\right]=\infty}, then the value of \mathrm{\text{a and b}} are
 

Option: 1

\mathrm{a=1, b \neq 1}
 


Option: 2

\mathrm{a \neq 1, b \neq 1}
 


Option: 3

\mathrm{a \neq 1, b\: \text{can have any value}}
 


Option: 4

\mathrm{a=1, b=1}


Answers (1)

best_answer

\mathrm{\operatorname{Lt}_{x \rightarrow \infty} \frac{x^2(1-a)-x(a+b)+(1-b)}{x+1}=\infty}
Above will be possible if degree of \mathrm{N^r} is greater than degree of \mathrm{D^r}

\mathrm{\therefore 1-a \neq 0\: i.e., a \neq 1\: and \: b} can have any value.

Hence option 3 is correct.

Posted by

sudhir kumar

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