$\underset{x→0}{\lim}\begin{bmatrix}\begin{pmatrix}3x+\frac{1}{x}\end{pmatrix}^2-\begin{pmatrix}2x+\frac{1}{x}\end{pmatrix}^2\end{bmatrix}$ is equal to
Answer & explanation
Correct answer: option 2
$\underset{x→0}{\lim}\begin{bmatrix}\begin{pmatrix}3x+\frac{1}{x}\end{pmatrix}^2-\begin{pmatrix}2x+\frac{1}{x}\end{pmatrix}^2\end{bmatrix}$
$=\underset{x→0}{\lim}\begin{bmatrix}\begin{pmatrix}3x+\frac{1}{x}+2x+\frac{1}{x}\end{pmatrix}\begin{pmatrix}3x+\frac{1}{x}-2x-\frac{1}{x}\end{pmatrix}\end{bmatrix}$
$=\underset{x→0}{\lim}\left[\left(5x+\frac{2}{x}\right)^{(x)}\right]=\underset{x→0}{\lim}(5x^2+2)$
$=2$