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Limits — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsLimits4 Marks
Question
Evaluate the following limit:
$\text{f(x)}=\frac{\text{ax}^2+\text{b}}{\text{x}^2+1},\lim\limits_{\text{x}\rightarrow0}\text{ f(x)}=1 $ and $\lim\limits_{\text{x}\rightarrow\infty}\text{f(x)}=1,$ then prove that $\text{f}(-2)=\text{f}(2)=1.$
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Answer

$\text{f(x)}=\frac{\text{ax}^2+\text{b}}{\text{x}^2+1}$
Also $\lim\limits_{\text{x}\rightarrow0}\text{ f(x)}=1\cdots{(\text{i})}$ [Given]
$\Rightarrow\ \lim\limits_{\text{x}\rightarrow0}\frac{\text{ax}^2+\text{b}}{\text{x}^2+1}=1$
$\Rightarrow\ \frac{\lim\limits_{\text{x}\rightarrow0}\text{ax}^2+\text{b}}{\lim\limits_{\text{x}\rightarrow0}\text{x}^2+1}=1$
$\Rightarrow\ \text{b}=1$
Also, it is given that $\lim\limits_{\text{x}\rightarrow\infty}\text{f(x)}=1$
$\therefore\ \lim\limits_{\text{x}\rightarrow\infty}\frac{\text{ax}^2+\text{b}}{\text{x}^2+1}=1\ \cdots(\text{ii})$
$\Rightarrow\ \lim\limits_{\text{x}\rightarrow\infty}\frac{\text{ax}^2+\text{b}}{\text{x}^2+1}=1$
$\Rightarrow\ \lim\limits_{\text{x}\rightarrow\infty}\frac{\text{ax}+\frac{1}{\text{x}^2}}{1+\frac{1}{\text{x}^2}}$ $\Big[\frac{\infty}{\infty}\text{ from}\Big]$
$\Rightarrow\text{ a}=1$
Thus, $\text{f(x)}=\frac{\text{ax}^2+\text{b}}{\text{x}^2+1}=\frac{\text{x}^2+1}{\text{x}^2+1}=1$ [From (ii)]
$\text{f}(-2)=1$
$\text{f}(2)=1$
$\text{f}(-2)=1=\text{f}(2)$
Hence, proved.

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