Why does $\frac{x-2}{|x-2|}$ make straight lines when x approaches 2 and not asymptotes to infinity?

[Integrate]

[math]\frac{x-2}{|x-2|}[/math]
As x approaches 2 from the left the denominator becomes infinitely big while the numerator becomes infinitely small but negative meaning the left handed limit goes to negative infinity.

[math]\lim_{x \to 2^-}\frac{x-2}{|x-2|}=-\infty[/math]The opposite is true when x approaches 2 from the right. We get a positive numerator which gives positive infinity.

[math]\lim_{x \to 2^+}\frac{x-2}{|x-2|}=\infty[/math]

However the graph gives a much different answer.




What have I done wrong?

 

[BigBeachBanana]

Your one-sided limits are incorrect. They are -1 and 1. How did you get infinity?

 

[Integrate]

Doesn't the denominator become infinitely large in each case? There fore taking them to infinity in their respective direction?

Your one-sided limits are incorrect. They are -1 and 1. How did you get infinity?

 

[BigBeachBanana]

Doesn't the denominator become infinitely large in each case? There fore taking them to infinity in their respective direction?

Did you consider the numerator?

 

[Integrate]

Did you consider the numerator?

The numerator becomes infinitely small yes. Does that mean that infinity over infinity can be expressed as the integer one?

 

[BigBeachBanana]

The numerator becomes infinitely small yes. Does that mean that infinity over infinity can be expressed as the integer one?

What do you mean by infinitely small? It's approaching 0, so you get 0/0, an indeterminant form. Did you learn l'Hopital? and it's not always 1.

 

[Dr.Peterson]

[math]\frac{x-2}{|x-2|}[/math]
As x approaches 2 from the left the denominator becomes infinitely big while the numerator becomes infinitely small but negative meaning the left handed limit goes to negative infinity.

[math]\lim_{x \to 2^-}\frac{x-2}{|x-2|}=-\infty[/math]The opposite is true when x approaches 2 from the right. We get a positive numerator which gives positive infinity.

[math]\lim_{x \to 2^+}\frac{x-2}{|x-2|}=\infty[/math]

However the graph gives a much different answer.

View attachment 31255


What have I done wrong?

When both numerator and denominator approach zero, it's indeterminate (0/0). That means you need to stop and think carefully!

For the limit from the left, for example, rewrite the expression and simplify: [math]\lim_{x \to 2^-}\frac{x-2}{|x-2|}=\lim_{x \to 2^-}\frac{x-2}{-(x-2)}=\lim_{x \to 2^-}(-1)=-1[/math]
Do the same on the right.

 

[Integrate]

When both numerator and denominator approach zero, it's indeterminate (0/0). That means you need to stop and think carefully!

For the limit from the left, for example, rewrite the expression and simplify: [math]\lim_{x \to 2^-}\frac{x-2}{|x-2|}=\lim_{x \to 2^-}\frac{x-2}{-(x-2)}=\lim_{x \to 2^-}(-1)=-1[/math]
Do the same on the right.

Thank you Dr. Peterson. Always a great help.

 

[Integrate]

What do you mean by infinitely small? It's approaching 0, so you get 0/0, an indeterminant form. Did you learn l'Hopital? and it's not always 1.

Thank you for your help. Dr. Peterson was able to take me across the finish line.

 

[Steven G]

Recall that |x-2| = x-2 or it equals -(x-2). Ignoring the sign of the fraction for a moment the numerator and denominator are the same! Regardless of what x is approaching how can the denominator get large and the numerator get small. Again, the numerator and denominator are the basically the same. If you explain what you were thinking someone here will clear things up for you.

 

[Integrate]

Recall that |x-2| = x-2 or it equals -(x-2). Ignoring the sign of the fraction for a moment the numerator and denominator are the same! Regardless of what x is approaching how can the denominator get large and the numerator get small. Again, the numerator and denominator are the basically the same. If you explain what you were thinking someone here will clear things up for you.

This made it click even further! So obvious now.

Thank you a ton!