Calculus: Limit and continuity 
Would anyone mind telling me how to solve these two questions? I know it sounds silly but I really have no idea.
 A: The limit of the first radical is obviously $0$, and $\cos$ is a bound function, whose values are always in between $\pm1$, so the limit of the product is $0$ as well. In the second case, amplify with the conjugate.
A: Hints:


*

*$\left|\cos\right|$ is bounded by 1

*Multiply by $\displaystyle \frac{-\sqrt{x^2+x}-x}{-\sqrt{x^2+x}-x}$

*Assume $\left|x\right|\lt1$

A: For 2) first limit is B and second limit is A. The reason is that when $x \to 0^{+}$ then $x^{3} - x = x(x^{2} - 1) \to 0^{-}$ and when $x \to 0^{-}$ then $x^{2}-x^{4} = x^{2}(1 - x^{2}) \to 0^{+}$.
Update: After some further clarification requested by OP, I elaborate a bit further. Note that as $x\to 0^{+}$ we see that $x$ is positive and small so that ultimately $0 < x < 1$ and hence $0 < x^{2} < 1$ and hence $(x^{2} - 1) < 0$ and therefore $x(x^{2} - 1) < 0$. But at the same time we need to note that $x(x^{2} - 1) \to 0\cdot (0 - 1) = 0$. It thus follows that the variable $y = x^{3} - x = x(x^{2} - 1)$ tends to zero but is negative so that $y \to 0^{-}$ and hence $\lim_{x \to 0^{+}}f(x^{3} - x) = \lim_{y \to 0^{-}}f(y) = B$. The other question can also be handled similarly.
