[tex]\displaystyle\lim_{x\to-2^+}\frac{x-1}{x^2(x+2)}[/tex]
The limit is infinite because the denominator approaches 0 while the numerator does not, since [tex]x+2=0[/tex] when [tex]x=-2[/tex]. Which infinity it approaches (positive or negative) depends on the sign of the other terms for values of [tex]x[/tex] near -2.
Since [tex]x\to-2[/tex] from the right, we're considering values of [tex]x>-2[/tex]. For example, if [tex]x=-1.9[/tex], then [tex]\dfrac{x-1}{x^2}=\dfrac{-2.9}{1.9^2}<0[/tex]; if [tex]x=-1.99[/tex], then [tex]\dfrac{x-1}{x^2}=\dfrac{-2.99}{1.99^2}<0[/tex], and so on. We can keep picking values of [tex]x[/tex] that get closer and closer to -2, and we would see that [tex]\dfrac{x-1}{x^2}[/tex] contributes a negative sign every time. So the limit must be [tex]\boxed{-\infty}[/tex].
[tex]\displaystyle\lim_{x\to0}\frac{x-1}{x^2(x+2)}[/tex]
By similar reasoning above, we see that [tex]\dfrac{x-1}{x+2}[/tex] contributes a negative sign regardless of which side we approach 0 from. [tex]x-1[/tex] is always negative and [tex]x+2[/tex] is always positive, so the net effect is a negative sign and the limit from either side is [tex]\boxed{-\infty}[/tex].
[tex]\displaystyle\lim_{x\to2\pi}x\csc x=\lim_{x\to2\pi}\frac x{\sin x}[/tex]
Direct substitution gives 0 in the denominator. For [tex]x>2\pi[/tex] we have [tex]\sin x>0[/tex], and for [tex]x<2\pi[/tex] we have [tex]\sin x<0[/tex]. Meanwhile, the numerator stays positive, which means the limit is positive or negative infinity depending on the direction in which [tex]x[/tex] approaches [tex]2\pi[/tex], so this limit does not exist.
