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Limits — MATHS STD 11 Science — Question

Gujarat 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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