Question

$\lim_{x \to 1} (\frac{1}{1-x} - \frac{3}{1-x^3})$ is equal to:

• A. -1
• B. 0
• C. 1
• D. D.N.E.

Answer 1

Answer: (A)

-1

Answer 2

To evaluate the limit $\lim_{x \to 1} (\frac{1}{1-x} - \frac{3}{1-x^3})$, we first combine the two fractions into a single one.

Step 1: Combine the fractions. The common denominator for $(1-x)$ and $(1-x^3)$ is $(1-x^3)$, since $1-x^3 = (1-x)(1+x+x^2)$. $\frac{1}{1-x} - \frac{3}{1-x^3} = \frac{1(1+x+x^2)}{(1-x)(1+x+x^2)} - \frac{3}{(1-x)(1+x+x^2)}$ $= \frac{1+x+x^2-3}{(1-x)(1+x+x^2)}$ $= \frac{x^2+x-2}{(1-x)(1+x+x^2)}$

Step 2: Factor the numerator. The numerator is a quadratic expression $x^2+x-2$. We look for two numbers that multiply to -2 and add to 1. These numbers are 2 and -1. So, $x^2+x-2 = (x+2)(x-1)$.

Step 3: Substitute the factored numerator and simplify. Substitute the factored numerator back into the expression: $\frac{(x+2)(x-1)}{(1-x)(1+x+x^2)}$ We notice that $(x-1) = -(1-x)$. Substitute this into the expression: $\frac{(x+2)(-(1-x))}{(1-x)(1+x+x^2)}$ For $x \neq 1$, we can cancel the term $(1-x)$ from the numerator and the denominator: $= \frac{-(x+2)}{1+x+x^2}$

Step 4: Evaluate the limit. Now, we can evaluate the limit by substituting $x=1$ into the simplified expression, as the denominator $1+x+x^2$ is not zero at $x=1$ (it becomes $1+1+1^2=3$). $\lim_{x \to 1} \frac{-(x+2)}{1+x+x^2} = \frac{-(1+2)}{1+1+1^2}$ $= \frac{-3}{3}$ $= -1$

Alternative Method: Using L'Hopital's Rule After combining the fractions, we got the form $\frac{x^2+x-2}{1-x^3}$. When $x=1$, the numerator is $1^2+1-2=0$ and the denominator is $1-1^3=0$. This is an indeterminate form $\frac{0}{0}$, so we can apply L'Hopital's Rule. Let $f(x) = x^2+x-2$ and $g(x) = 1-x^3$. Then $f'(x) = 2x+1$ and $g'(x) = -3x^2$. $\lim_{x \to 1} \frac{f'(x)}{g'(x)} = \lim_{x \to 1} \frac{2x+1}{-3x^2}$ Substitute $x=1$: $= \frac{2(1)+1}{-3(1)^2} = \frac{3}{-3} = -1$

Both methods yield the same result.