How do you know a limit is towards infinity or is undefined

Question

$\lim_{x\to1}\dfrac{x-1}{x^2+x-2}$ = $\dfrac{1}{3}$, but

$\lim_{x\to-2+}\dfrac{x-1}{x^2+x-2}$ tends towards positive infinity and

$\lim_{x\to-2-}\dfrac{x-1}{x^2+x-2}$ tends towards negative infinity.

How do you know this without the need of trying different values of x very close to where the limit tends to?

Answer 1

First of all,

$$\lim_{x\to1} \frac{x-1}{x^2+x-2} = \lim_{x \to 1} \frac{x-1}{(x-1)(x+2)} = \frac{1}{3}.$$

To see what a limit is, we have to try values that are "close" to the limiting point. When you do this enough, your intuition allows you to understand what it should tend to. For example, with your second one, the numerator is always positive and $> 1$, and the denominator tends to $0$, so it "blows up" to positive infinity. Of course, this is just a heuristic, and a good rigorous treatment is always needed when you encounter something you are unfamiliar with.

Answer 2

if $x\ne 1$ or $-2$ then $\dfrac{x-1}{x^2+x-2}= \dfrac {x-1}{(x-1)(x+2)} = \dfrac{1}{x+2}$

So

$\lim_{x\to1}\dfrac{x-1}{x^2+x-2}=\lim_{x\to 1} \frac 1{x+2} = \frac 13$ (not undefined at all; the expression is when evaluated at $x=1$ but the limit of the expression is not undefined).

$\lim_{x\to 2^-}\dfrac{x-1}{x^2+x-2}=\lim_{x\to 2^-} \frac 1{x+2}=\lim_{y-2\to 0^-}\frac 1y = -\infty$

$\lim_{x\to 2^+}\dfrac{x-1}{x^2+x-2}=\lim_{x\to 2^+} \frac 1{x+2}=\lim_{y-2\to 0^+}\frac 1y = \infty$

Answer 3

First of all notice that $$\dfrac{x-1}{x^2+x-2}= \dfrac {x-1}{(x-1)(x+2)} = \dfrac{1}{x+2}$$

If you let $y=x+2$, you fraction is simply $1/y$ , where you want to find the limit as $y\to 0$

As you know $y$ and $1/y$ have the same sign.

Thus as $y\to 0$ the right limit of $1/y$ is $+\infty $ while the left limit is $-\infty $

Answer 4

at first we simplify fraction to

$$\frac{x-1}{x^2+x-2} = \frac{x-1}{(x-1)(x+2)} = \frac{1}{x+2}.$$

2 is root of Denominator and if you placement $2^-$ in fraction we have a fraction like this

$$\frac{1}{0^-}$$ and it's equall to $-\infty$

and also we placement $2^+$ in fraction we have a fraction like this

$$\frac{1}{0^+}$$ and it's equall to $\infty$