Mathematics - Limits Question with Solution | TestHub

$\mathrm{L}={\mathrm{Lim}}_{\mathrm{x}\to 0}\frac{\sin \left({\mathrm{x}}^2\right)+2\cos (\mathrm{bx})-{\mathrm{ax}}^4-2}{{\mathrm{e}}^{\mathrm{ax}}-1-\mathrm{ax}-2{\mathrm{x}}^2-\frac{{\mathrm{a}}^3{\mathrm{x}}^3}{6}}$

$={\mathrm{Lim}}_{x\to 0}\frac{x^2-\frac{x^6}{3!}+\dots .+2\left(1-\frac{b^2{x}^2}{2!}+\frac{b^4{x}^4}{4!}+\dots .\right)-a{x}^4-2}{1+ax+\frac{a^2{x}^2}{2!}+\frac{a^3{x}^3}{3!}+\frac{a^4{x}^4}{4!}+\dots .-1-ax-2x^3-\frac{a^3{x}^3}{6}}$

$={\mathrm{Lim}}_{x\to 0}\frac{\left(1-b^2\right)x^2+\left(\frac{b^2}{12}-a\right)x^4+\dots \dots }{\left(\frac{a^2}{2}-2\right)x^2+\frac{a^4{x}^4}{24}+\dots }$

$\frac{1-{\mathrm{b}}^2}{\frac{{\mathrm{a}}^2}{2}-2}=\frac{25}{8}\Rightarrow 16-16{\mathrm{b}}^2=25\left({\mathrm{a}}^2-4\right)$

$\Rightarrow 25{\mathrm{a}}^2+16{\mathrm{b}}^2=116$

$\text{displaylines}\Rightarrow \begin{array}{c}{a}^2=4;b^2=1\\ a=\pm 2,b=\pm 1\end{array}$ (∵ $\left(a,b\in I\right)$

$\frac{1-{\mathrm{b}}^2}{\frac{{\mathrm{a}}^2}{2}-2}=\frac{-23}{8}\Rightarrow 16-16{\mathrm{b}}^2=-23\left({\mathrm{a}}^2-4\right)$

$23{\mathrm{a}}^2-16{\mathrm{b}}^2=76\Rightarrow {\mathrm{a}}^2=4$

$\Rightarrow \begin{array}{c}{a}^2=4;b^2=1\\ a=\pm 2,b=\pm 1\\ (\because a,b\in I)\end{array}$

Now, $L=\frac{\frac{b^2}{12}-a}{\frac{a^4}{24}}=\frac{2\left(b^2-12a\right)}{a^4}$

For $\mathrm{a}=2,\mathrm{b}=\pm 1$

$\Rightarrow \mathrm{L}=\frac{2\cdot (1-24)}{16}=\frac{-23}{8}$

For $\mathrm{a}=-2,\mathrm{b}=\pm 1$

$\Rightarrow \mathrm{L}=\frac{2\cdot (1+24)}{16}=\frac{25}{8}$