Question

Lim((x^3+1)/(x^2-1) as x->-1)

Answer 1

$\lim\limits_{x \to -1}\frac{x^3+1}{x^2-1}=\frac{(-1)^3+1}{(-1)^2-1}=\frac{-1+1}{1-1}=\frac00=\\ \|x^3+1=(x+1)(x^2-x+1)=x^3-x^2+x+x^2-x+1=x^3+1\|\\ \|x^2-1=(x+1)(x-1) \|\\ =\lim\limits_{x \to -1}\frac{(x+1)(x^2-x+1)}{(x-1)(x+1)}=\lim\limits_{x \to -1}\frac{(x^2-x+1)}{(x-1)}=\frac{(-1)^2-(-1)+1}{(-1)-1}=\\ =\frac{1+1+1}{-1-1}=\frac3{-2}=-\frac32=-1\frac12.$
или о правилу лопиталя
$\lim\limits_{x \to -1}\frac{x^3+1}{x^2-1}=\frac{0}{0}\stackrel{\Lambda}{=}\lim\limits_{x \to -1}\frac{\left(x^3+1\right)'}{\left(x^2-1\right)'}=\frac{3x^2}{2x}=\frac{3\cdot(-1)^2}{2\cdot(-1)}=\\ \\ \\ =\frac{3\cdot1}{-2}=\frac{3}{-2}=-\frac32=-1\frac12.$

Answer 2

To evaluate the limit:

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lim   (x^3 + 1)
x → -1  (x^2 - 1)

We can directly substitute x = -1 into the expression to see what happens:

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(-1)^3 + 1
lim   ------------ = 0/0  (indeterminate form)
x → -1  (-1)^2 - 1

We have obtained an indeterminate form of 0/0, which means we cannot directly evaluate the limit by substitution.

To proceed, we can factor the numerator and denominator and simplify:

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x^3 + 1 = (x + 1)(x^2 - x + 1)
x^2 - 1 = (x + 1)(x - 1)

Substituting these expressions back into the original limit, we get:

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lim   (x + 1)(x^2 - x + 1)
x → -1  ------------------
            (x + 1)(x - 1)

We can cancel out the common factor of (x + 1) in both the numerator and denominator:

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lim   (x^2 - x + 1)
x → -1  -------------
            (x - 1)

Now we can substitute x = -1 into the expression and evaluate the limit:

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(-1)^2 - (-1) + 1     3
lim   -------------- = --- = 3
x → -1  (-1) - 1      -2

Therefore, the limit of (x^3+1)/(x^2-1) as x approaches -1 is equal to 3.