Search any question & find its solution
Question:
Answered & Verified by Expert
$\lim _{x \rightarrow \infty}\left(\frac{x^2+1}{x+1}-a x-b\right),(a, b \in R)=0$. Then
Options:
Solution:
1652 Upvotes
Verified Answer
The correct answer is:
$a=1, b=-1$
$\lim _{x \rightarrow \infty} \frac{x^2+1-a x(x+1)-b(x+1)}{x+1}=0$
$=\lim _{x \rightarrow \infty} \frac{(1-a) x^2-(a+b) x+1-b}{x+1}=0$
for the limit to exist,
$\begin{aligned}
1-\mathrm{a} &=0 \text { and }-(\mathrm{a}+\mathrm{b})=0 \\
\therefore \mathrm{a}=1, \mathrm{~b} &=-1
\end{aligned}$
$=\lim _{x \rightarrow \infty} \frac{(1-a) x^2-(a+b) x+1-b}{x+1}=0$
for the limit to exist,
$\begin{aligned}
1-\mathrm{a} &=0 \text { and }-(\mathrm{a}+\mathrm{b})=0 \\
\therefore \mathrm{a}=1, \mathrm{~b} &=-1
\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.