Mathematics - Limits Question with Solution | TestHub

$\begin{array}{rl}& \underset{x\to \infty }{\lim }\frac{\sin x^2\cdot {\left({\log }_{e}x\right)}^{\alpha }\cdot \sin \frac{1}{x^2}}{x^{\alpha \beta }\cdot {\left({\log }_e(1+x)\right)}^{\beta }}=0\\ & \underset{x\to \infty }{\lim }\frac{{\left({\log }_{e}x\right)}^{\alpha }}{{\left({\log }_e(x+1)\right)}^{\beta }\cdot x^{\alpha \beta +2}}=0\end{array}$ $\begin{array}{rl}& \underset{x\to \infty }{\lim }{\left(\frac{{\log }_{e}x}{{\log }_e(x+1)}\right)}^{\beta }\cdot \frac{{\left({\log }_{e}x\right)}^{\alpha -\beta }}{x^{\alpha \beta +2}}=0\\ & \underset{x\to \infty }{\lim }\frac{{\left({\log }_{e}x\right)}^{\alpha -\beta }}{x^{\alpha \beta +2}}=0\text{ Put }{\log }_{e}x=t\\ & \underset{t\to \infty }{\lim }\frac{t^{\alpha -\beta }}{{\left(e^t\right)}^{\alpha \beta +2}}=0\end{array}$ As we know ${\lim }_{x\to \infty }\frac{x}{e^x}=0$ $\alpha \beta +2>0\Rightarrow \alpha \beta >-2$