find the limit of y = $\frac{x^3 + 6x - 7}{x-1}$ as x tends to 1

1. Analyze the question

We are asked to find the limit of the function $f(x) = \frac{x^3 + 6x - 7}{x-1}$ as $x \to 1$.

If we substitute $x = 1$ directly into the expression, we get: $\frac{1^3 + 6(1) - 7}{1 - 1} = \frac{1 + 6 - 7}{0} = \frac{0}{0}$

This is an indeterminate form. In calculus, when you encounter $\frac{0}{0}$, it means there is a common factor in the numerator and denominator that can be cancelled out.

2. Explain the concept

To solve limits that result in $\frac{0}{0}$, we can use two main methods:

• Factorization: Simplify the expression by factoring the numerator.
• L'Hôpital's Rule: Differentiate the numerator and the denominator separately with respect to $x$ and then substitute the limit value.

3. Show the solution

Method 1: Factorization Since $(x-1)$ is a factor of the denominator, it must also be a factor of the numerator. We can perform polynomial long division or synthetic division to divide $(x^3 + 6x - 7)$ by $(x-1)$.

Dividing $x^3 + 0x^2 + 6x - 7$ by $(x-1)$:

• $x^3 \div x = x^2$
• $x^2(x-1) = x^3 - x^2$
• Subtracting: $(x^3 + 0x^2) - (x^3 - x^2) = x^2$
• Bring down $6x$: $x^2 + 6x$
• $x^2 \div x = x$
• $x(x-1) = x^2 - x$
• Subtracting: $(x^2 + 6x) - (x^2 - x) = 7x$
• Bring down $-7$: $7x - 7$
• $7x \div x = 7$
• $7(x-1) = 7x - 7$
• Remainder = 0

So, $x^3 + 6x - 7 = (x-1)(x^2 + x + 7)$.

Now, substitute this back into the limit: $\lim_{x \to 1} \frac{(x-1)(x^2 + x + 7)}{(x-1)}$

Cancel the $(x-1)$ terms: $\lim_{x \to 1} (x^2 + x + 7)$

Substitute $x = 1$: $1^2 + 1 + 7 = 1 + 1 + 7 = 9$

Method 2: L'Hôpital's Rule Differentiate the numerator and denominator: Numerator: $\frac{d}{dx}(x^3 + 6x - 7) = 3x^2 + 6$ Denominator: $\frac{d}{dx}(x - 1) = 1$

Now find the limit of the new fraction: $\lim_{x \to 1} \frac{3x^2 + 6}{1} = 3(1)^2 + 6 = 3 + 6 = 9$

4. Identify the answer

Both methods confirm that the limit is 9. This corresponds to option A.

5. Tip

Whenever you see a limit problem where the denominator becomes zero, always check for the $\frac{0}{0}$ case. If you see it, don't panic! Just factorize the numerator or use L'Hôpital's Rule (if you are comfortable with differentiation). It is a guaranteed way to clear the "undefined" part of the fraction.