$\lim_{x o 3}\dfrac{x^4 - 81}{2x^2 - 5x - 3}$

Question

$\lim_{x	o 3}\dfrac{x^4 - 81}{2x^2 - 5x - 3}$

Accepted Answer

> [!TLDR]
> The limit is of the form $\frac{0}{0}$ because both numerator and denominator vanish at $x=3$. Factorising the numerator as a difference of squares and the denominator as a quadratic, we cancel the common factor $(x-3)$ and then substitute $x=3$ to get the value $\frac{108}{7}$.

The first thing to notice is that if you plug $x=3$ directly into the expression, you get $\frac{81-81}{18-15-3} = \frac{0}{0}$. That’s an indeterminate form — it doesn’t tell us the limit yet. But it does tell us that both the numerator and denominator have a factor of $(x-3)$.

Why? Because a polynomial $P(x)$ has $P(3)=0$ if and only if $(x-3)$ is a factor. So the $\frac{0}{0}$ form is a signal: factor and cancel.

> [!TIP]
> For any polynomial limit that gives $\frac{0}{0}$, always check if the numerator and denominator share a common root. If they do, factorisation (or synthetic division) will remove the indeterminacy.

Let’s work through it step by step.

1. **Factor the numerator**  
   $x^4 - 81$ is a difference of squares:  
   $$x^4 - 81 = (x^2)^2 - 9^2 = (x^2 - 9)(x^2 + 9)$$  
   And $x^2 - 9$ is itself a difference of squares:  
   $$x^2 - 9 = (x-3)(x+3)$$  
   So the numerator becomes:  
   $$x^4 - 81 = (x-3)(x+3)(x^2+9)$$

2. **Factor the denominator**  
   $2x^2 - 5x - 3$ is a quadratic. We look for two numbers that multiply to $2 	imes (-3) = -6$ and add to $-5$. Those numbers are $-6$ and $1$. So:  
   $$2x^2 - 5x - 3 = 2x^2 - 6x + x - 3 = 2x(x-3) + 1(x-3) = (x-3)(2x+1)$$

3. **Cancel the common factor** …

Q. $\lim_{x\to 3}\dfrac{x^4 - 81}{2x^2 - 5x - 3}$

Concept understanding — Algebra of Limits

Q.lim⁡x→3x4−812x2−5x−3\lim_{x\to 3}\dfrac{x^4 - 81}{2x^2 - 5x - 3}limx→3​2x2−5x−3x4−81​

🎓Unlock everything free for 14 days

Q. $\lim_{x\to 3}\dfrac{x^4 - 81}{2x^2 - 5x - 3}$

Unlock everything free for 14 days