How to evaluate a rational limit by factoring and canceling

Expert intro: This limit looks messy at first, but the structure is friendly once you notice the common factor. The trick is to simplify before substituting, because direct substitution would give you $0/0$ and stop the computation cold.

Detailed walkthrough

We want to evaluate

$\lim_{x\to 0}\frac{x^2+3x}{3x^2-7x}$

Step 1: Factor numerator and denominator

$x^2+3x=x(x+3)$

$3x^2-7x=x(3x-7)$

So the limit becomes

$\lim_{x\to 0}\frac{x(x+3)}{x(3x-7)}$

Step 2: Cancel the common factor

For $x\neq 0$, we can cancel $x$:

$\lim_{x\to 0}\frac{x+3}{3x-7}$

Step 3: Substitute $x=0$

$\frac{0+3}{3(0)-7}=\frac{3}{-7}=-\frac{3}{7}$

Answer

$\boxed{-\frac{3}{7}}$

Quick check

If you substitute directly into the original expression, both numerator and denominator become 0, so factoring is the correct move here.

💡 Pitfall guide

The main trap is copying the sign incorrectly after canceling. In the denominator, $3x-7$ is not $3x+7$. Another easy miss is forgetting that the expression after cancellation is only valid for $x\neq 0$, which is fine here because we are taking a limit, not evaluating at $x=0$ directly.

🔄 Real-world variant

If the denominator had been $3x^2+7x$ instead, the same factoring idea would give

$\lim_{x\to 0}\frac{x(x+3)}{x(3x+7)}=\lim_{x\to 0}\frac{x+3}{3x+7}=\frac{3}{7}$

So a small sign change can flip the final answer. That is why it helps to rewrite both factors carefully before simplifying.

🔍 Related terms

factoring rational expressions, indeterminate form, limit laws