The value of $\underset{x→∞}{\lim}\begin{pmatrix}\frac{3x^2+1}{4x^2-1}\end{pmatrix}^{\frac{x^3}{1+x}}$ is

0

$\underset{x→∞}{\lim}\begin{bmatrix}\frac{3x^2+1}{4x^2-1}\end{bmatrix}^{\frac{x^3}{1+x}}=\underset{x→∞}{\lim}\begin{bmatrix}\frac{3(x^2+\frac{1}{3})}{4(x^2-\frac{1}{4})}\end{bmatrix}^{\frac{x^3}{1+x}}$

$=\underset{x→∞}{\lim}(\frac{3}{4})^{\frac{x^3}{1+x}}\begin{bmatrix}\frac{x^2+\frac{1}{3}}{x^2-\frac{1}{4}}\end{bmatrix}^{\frac{x^3}{1+x}}=\underset{x→∞}{\lim}(\frac{3}{4})^{\frac{x^3}{1+x}}\begin{bmatrix}1+\frac{\frac{7}{12}}{x^2-\frac{1}{4}}\end{bmatrix}^{\frac{x^3}{1+x}}=(\frac{3}{4})^{\underset{x→∞}{\lim}\frac{x^3}{1+x}}.\underset{x→∞}{\lim}\begin{bmatrix}1+\frac{\frac{7}{12}}{x^2-\frac{1}{4}}\end{bmatrix}^{\frac{x^3}{1+x}}$

Now $\underset{x→∞}{\lim}\frac{x^3}{1+x}=\underset{x→∞}{\lim}\frac{x^2}{\frac{1}{x}+1}=∞$

$(\frac{3}{4})^{\underset{x→∞}{\lim}\frac{x^3}{1+x}}=(\frac{3}{4})^∞=0\,\,(∵0<\frac{3}{4}<1)$

and $\underset{x→∞}{\lim}\begin{bmatrix}1+\frac{\frac{7}{12}}{x^2-\frac{1}{4}}\end{bmatrix}^{\frac{x^3}{1+x}}=\underset{x→∞}{\lim}\begin{bmatrix}\begin{Bmatrix}1+\frac{\frac{7}{12}}{x^2-\frac{1}{4}}\end{Bmatrix}^{x^2-\frac{1}{4}}\end{bmatrix}^{\frac{x^3}{(x^2-\frac{1}{4})(x+1)}}$

$=(e^{\frac{7}{12}})^1=e^{\frac{7}{12}}$  $\begin{bmatrix}∵\underset{x→∞}{\lim}\begin{bmatrix}1+\frac{a}{f(x)}\end{bmatrix}^{f(x)}\end{bmatrix}=e^a$

and $\underset{x→∞}{\lim}=\frac{x^3}{(x^2+1/x)(1+x)}=\underset{x→∞}{\lim}\frac{1}{(1-1/4x^2)(1/4+1)}=1$

∴ The required limit = $0.e^{\frac{7}{12}}=0$