Find if the limit is +ve infinity or -ve infinity

1)$\displaystyle lim (x \rightarrow 0^-)\frac{\sqrt{x^2+3}}{x} $

2)$\displaystyle lim (x \rightarrow 0^+)\frac{1}{x}-\frac{1}{x^2} $

Attempt:

1)$\displaystyle -\infty $

2)$\displaystyle +\infty $

Am I correct?