Question

$\underset{x\to \infty }{\lim }{\left\{\frac{3x^2+1}{4x^2-1}\right\}}^{\frac{x^3}{1+x}}$

Answer 1

$$\begin{gathered} \underset{x\to \infty }{\lim }{\left\{\frac{3x^2+1}{4x^2-1}\right\}}^{\frac{x^3}{1+x}} \\ =\underset{x\to \infty }{\lim }{\left\{1+\frac{3x^2+1}{4x^2-1}-1\right\}}^{\frac{x^3}{1+x}} \\ =\underset{x\to \infty }{\lim }{\left\{1+\frac{3x^2+1-4x^2+1}{4x^2-1}\right\}}^{\frac{x^3}{1+x}} \\ =\underset{x\to \infty }{\lim }{\left\{1+\frac{2-x^2}{4x^2-1}\right\}}^{\frac{x^3}{1+x}} \\ =e^{\underset{x\to \infty }{\lim }\left(\frac{2-x^2}{4x^2-1}\right)\times \left(\frac{x^3}{1+x}\right)} \\ =e{}^{\underset{x\to \infty }{\lim }\left(\frac{-x^5+2x^3}{\left(4x^2-1\right)\left(1+x\right)}\right)} \\ \mathrm{Dividing}{\mathrm{N}}^r\mathrm{and}{\mathrm{D}}^r\mathrm{by}{x}^5: \\ =e{}^{\underset{x\to \infty }{\lim }\left(\frac{-1+{\displaystyle \frac{2}{x^3}}}{{\displaystyle \frac{\left(4x^2-1\right)}{x^3}}{\displaystyle \frac{\left(1+x\right)}{x^2}}}\right)} \\ =e{}^{\underset{x\to \infty }{\lim }\left(\frac{-1+{\displaystyle \frac{2}{x^3}}}{\left({\displaystyle \frac{4}{x}}-{\displaystyle \frac{1}{x^3}}\right)\left({\displaystyle \frac{1}{x^2}}+{\displaystyle \frac{1}{x}}\right)}\right)} \\ =e^{-\frac{1}{0}} \\ =e^{-\infty }=0 \end{gathered}$$