Limits confused

[Loki123]

Is this correct?
I am supposed to get negative infinity, which I did get. If it's correct why is diving less than 6 by less than 0, negative infinity??

 

[BigBeachBanana]

$$\lim_{x \to 4^-}\frac{x^2-3x+2}{x-4}=\left(\lim_{x \to 4^-}x^2-3x+2 \right)\cdot \left(\lim_{x \to 4^-}\frac{1}{x-4}\right)=6(-\infty)=-\infty$$

 

[Loki123]

$$\lim_{x \to 4^-}\frac{x^2-3x+2}{x-4}=\left(\lim_{x \to 4^-}x^2-3x+2 \right)\cdot \left(\lim_{x \to 4^-}\frac{1}{x-4}\right)=6(-\infty)=-\infty$$

How did you not get less than 6? I am confused.

 

[BigBeachBanana]

How did you not get less than 6? I am confused.

By the existence of limit theorem:
$$\text{If } \lim_{x \to 4} x^2-3x+2=6 \text{, then} \lim_{\red{x \to 4^-}} x^2-3x+2= \lim_{\red{x \to 4^+}} x^2-3x+2=6$$

 

[Loki123]

By the existence of limit theorem:
$$\text{If } \lim_{x \to 4} x^2-3x+2=6 \text{, then} \lim_{\red{x \to 4^-}} x^2-3x+2= \lim_{\red{x \to 4^+}} x^2-3x+2=6$$

Still don't get it. Look at my work, that's how I assume you are supposed to solve it.

 

[BigBeachBanana]

Still don't get it. Look at my work, that's how I assume you are supposed to solve it.

The two-side limit exists if the LHS and the RHS limits are equal. Did you learn this?
Since the two-side limit of $x^2-3x+2=6$ exists, that implies the LHS and the RHS limits are equal to 6. So
$$\lim_{x \to 4^-} x^2-3x+2= \lim_{x \to 4^+} x^2-3x+2=\lim_{x \to 4} x^2-3x+2=6$$Go back to post#2:
$$\lim_{x \to 4^-}\frac{x^2-3x+2}{x-4}= \left(\lim_{x \to \red{4^-}}x^2-3x+2 \right)\cdot \left(\lim_{x \to 4^-}\frac{1}{x-4}\right)= \left(\lim_{x \to \red{4}}x^2-3x+2 \right) \cdot \left(\lim_{x \to 4^-}\frac{1}{x-4}\right)= 6(-\infty) =-\infty$$

 

[lex]

why is diving less than 6 by less than 0, negative infinity??

$x^2-3x+2 \to 6$ as $x \to4^-$

Eventually the top line of $\tfrac{x^2-3x+2}{x-4}$ is always positive (and $\to$ 6) as $x\to4^-$

Eventually the bottom line is always negative and becomes arbitrarily small as $x\to4^-$

$\tfrac{\text{positive} (\to 6)}{\text{small negative}}$ is large and negative, becoming arbitrarily large eventually as $x\to4^-$

 

[Steven G]

There is no need to say 6^-, which means a small amount less than 6. The reason is that any number around 6 is still positive. However, if a number is a little less than 0, ie 0^-, it IS negative and if a number is a little more than 0, ie 0⁺, then it IS positive. That is, there is a sign change around 0.

Now what does a little less than or a little more than mean? Let's look at the number a. The sign of a^- is the sign of $\displaystyle lim_ {n->\infty} (a - 1/n)$. There is a similar definition for a⁺