Question

Can anyone help me determine the infinite limit of these two problems

lim x--> -2^+(from the right)



`(x-1)/(x^(2) (x+2))`


lim x-->0 (from both left and right)



`(x-1)/(x^(2) (x+2))`

lim x-->2π⁻

x csc x


* I used direct substitution to determine the infinite limit of each but I have problems justifying how I got that answer by picking random points to plug into the equation

Answer 1

`\displaystyle\lim_(x\to-2^+)(x-1)/(x^2(x+2))`

The limit is infinite because the denominator approaches 0 while the numerator does not, since
`x+2=0` when
`x=-2`. Which infinity it approaches (positive or negative) depends on the sign of the other terms for values of
`x` near -2.

Since
`x\to-2` from the right, we're considering values of
`x>-2`. For example, if
`x=-1.9`, then
`(x-1)/(x^2)=(-2.9)/(1.9^2)<0`; if
`x=-1.99`, then
`(x-1)/(x^2)=(-2.99)/(1.99^2)<0`, and so on. We can keep picking values of
`x` that get closer and closer to -2, and we would see that
`(x-1)/(x^2)` contributes a negative sign every time. So the limit must be
`\boxed{-\infty}`.

`\displaystyle\lim_(x\to0)(x-1)/(x^2(x+2))`

By similar reasoning above, we see that
`(x-1)/(x+2)` contributes a negative sign regardless of which side we approach 0 from.
`x-1` is always negative and
`x+2` is always positive, so the net effect is a negative sign and the limit from either side is
`\boxed{-\infty}`.

`\displaystyle\lim_(x\to2\pi)x\csc x=\lim_(x\to2\pi)\frac x{\sin x}`

Direct substitution gives 0 in the denominator. For
`x>2\pi` we have
`\sin x>0`, and for
`x<2\pi` we have
`\sin x<0`. Meanwhile, the numerator stays positive, which means the limit is positive or negative infinity depending on the direction in which
`x` approaches
`2\pi`, so this limit does not exist.