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  • <img alt="\text{If}\lim _{x \rightarrow \infty}\left(\frac{x^{2}+1}{(x+1)}-a x-b\right)=0,\text{find\; a and b}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Ctext%7BIf%7D%5Clim%20_%7Bx%2

\text{If}\lim _{x \rightarrow \infty}\left(\frac{x^{2}+1}{(x+1)}-a x-b\right)=0,\text{find\; a and b}.

Option: 1

a=1, b=1


Option: 2

a=1, b=-1


Option: 3

a=-1, b=-1


Option: 4

a=-1$, $b=1


Answers (1)

best_answer

\lim _{x \rightarrow \infty}\left(\frac{x^{2}+1}{(x+1)}-a x-b\right)
=\lim _{x \rightarrow \infty} \frac{\left(x^{2}+1-a x^{2}-b x-a x-b\right)}{(x+1)}
=\lim _{x \rightarrow \infty} \frac{\left[(1-a) x^{2}-(a+b) x+1-b\right]}{(1+x)}=0 \text { (given) }

For this limit to exist, coefficient of  x^{2} must be zero.

1-a=0 \Rightarrow a=1
=\lim _{x \rightarrow \infty} \frac{[0-(a+b) x+1-b]}{(1+x)}=\lim _{x \rightarrow \infty} \frac{-(a+b)+\frac{1-b}{x}}{\frac{1}{x}+1}
=-(a+b)=0 \text { (given) }

\Rightarrow \quad b=-a \text { so, } b=-1

Posted by

Deependra Verma

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