Find: $\lim_{x o 8} \dfrac{2z^2 - 17z + 8}{8 - z}$.

Question

Find: $\lim_{x 	o 8} \dfrac{2z^2 - 17z + 8}{8 - z}$.

Accepted Answer

> [!TLDR]
> Direct substitution gives the indeterminate form $	frac{0}{0}$, so factor the numerator, cancel the common factor $(8-z)$, then substitute.

> [!FORMULA]
> **Factor-and-cancel rule:** if $\lim_{x	o a}\dfrac{f(x)}{g(x)}$ gives $	frac{0}{0}$ on direct substitution, factor out the common root $(x-a)$ from numerator and denominator, cancel it (valid since $x
e a$ while taking the limit), then substitute.

1. **Try direct substitution at $z=8$.**
   $$	ext{Numerator: } 2(8)^2-17(8)+8 = 128-136+8 = 0$$
   $$	ext{Denominator: } 8-8 = 0$$
   This is the indeterminate form $	frac{0}{0}$, so $(z-8)$ (equivalently $(8-z)$) must be a common factor.

2. **Factor the numerator** $2z^2-17z+8$. Look for two numbers whose product is $2	imes 8=16$ and whose sum is $-17$: these are $-16$ and $-1$.
   $$2z^2-17z+8 = 2z^2-16z-z+8 = 2z(z-8)-1(z-8) = (2z-1)(z-8)$$

3. **Rewrite the fraction**, noting $(z-8) = -(8-z)$. …

Q.Find: $\lim_{x \to 8} \dfrac{2z^2 - 17z + 8}{8 - z}$ .

Concept understanding — Calculus

What is Calculus?

The Intuition: Zooming In

The Precise Statement

Differential Calculus: The Derivative

Integral Calculus: The Integral

Q.Find: lim⁡x→82z2−17z+88−z\lim_{x \to 8} \dfrac{2z^2 - 17z + 8}{8 - z}limx→8​8−z2z2−17z+8​.

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Q.Find: $\lim_{x \to 8} \dfrac{2z^2 - 17z + 8}{8 - z}$ .

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