Does lim x→ -1 (2x^2+x-1)/(x^3+3x^2+5x+3)=0/0 ? If yes, then why does this algorithm give me 3/2 ? Am I crazy?

[Koalanet21]

I used the limit calculator from allmath.com https://www.allmath.com/limit-calculator.php
I entered (2x^2+x-1)/(x^3+3x^2+5x+3) because I didn't understand how to simplify (2x-1) and (x-1). But then the algorithm gave me "3/2". I was quite surprised so I recalculated one, two, three times... no matter how many times I tried, if I do it myself, I get 0/0, not 3/2.
I really don't understand. Is the algorithm broken or am I crazy?

Here's the doubtful answer it gave me.

 

[skeeter]

actually, you should get [MATH]-\frac{3}{2}[/MATH] as the limit

[MATH]\frac{2x^2+x-1}{x^3+3x^2+5x+3} = \frac{\cancel{(x+1)}(2x-1)}{\cancel{(x+1)}(x^2+2x+3)}[/MATH]
try it again

 

[Dr.Peterson]

Does lim x→ -1 (2x^2+x-1)/(x^3+3x^2+5x+3)=0/0 ? If yes, then why does this algorithm give me 3/2 ? Am I crazy?

I used the limit calculator from allmath.com https://www.allmath.com/limit-calculator.php
I entered (2x^2+x-1)/(x^3+3x^2+5x+3) because I didn't understand how to simplify (2x-1) and (x-1). But then the algorithm gave me "3/2". I was quite surprised so I recalculated one, two, three times... no matter how many times I tried, if I do it myself, I get 0/0, not 3/2.
I really don't understand. Is the algorithm broken or am I crazy?

Actually, it gives you -3/2, which is the correct answer.

Your answer of 0/0 is not an answer at all, as it is not a number. And your method appears not to involve finding a limit at all, but just direct substitution. That doesn't work in this case( which is called "indeterminate"). You need to do something else (such as skeeter's cancellation of common factors) to find the limit.

Can you explain your mention of (x - 1)? Possibly you factored incorrectly.

When you say "the algorithm", do you mean the program on that site? It's working fine.

 

[JeffM]

You have forgotten what the limit means.

[MATH]\exists \ \delta \text { such that } 0 < |x - a| < \delta \implies |f(x) - L| < \epsilon \iff \\ \displaystyle \lim_{x \rightarrow a} f(x) = L.[/MATH]In other words, you never, ever let x = - 1 if you are trying to find the limit at - 1.

So follow skeeter

[MATH]x \ne - 1 \implies f(x) = \dfrac{2x - 1}{x^2 + 2x + 3}.[/MATH]
Now there is no problem finding the limit of the RHS at x = - 1 to be - 3/2. Therefore the limit of the LHS is also - 3/2 because with limits you are never concerned with f(-1).

 

[Koalanet21]

Nevermind I just realized I'm inattentive enough not to click on "show steps" then "show step by step solution". Sorry for the inconvenience. You helped me by confirming me that "-3/2" was indeed the answer

 

[Steven G]

If x=-1 makes a polynomial =0, then x-(-1) = x+1 must be a factor. You do realize that the prerequisite for calculus 1 is precalculus?

 

[HallsofIvy]

You should understand that a limit is never "0/0". A limit, if it exists is a number and 0/0 is not a number!