Find the limit of a rational expression as x goes to negative infinity

Expert intro: At infinity, the highest power usually decides everything. Here the numerator and denominator grow at different rates, so the fraction shrinks fast.

Detailed walkthrough

We need to evaluate $\lim_{x\to -\infty}\frac{x-2}{x^2+2x+1}.$

Step 1: Compare degrees

The numerator has degree 1, and the denominator has degree 2. Since the denominator grows faster, the fraction should approach 0.

Step 2: Make it algebraic

Divide numerator and denominator by $x^2$:

=\frac{\frac{x}{x^2}-\frac{2}{x^2}}{1+\frac{2}{x}+\frac{1}{x^2}}$$ $$=\frac{\frac{1}{x}-\frac{2}{x^2}}{1+\frac{2}{x}+\frac{1}{x^2}}$$ ### Step 3: Take the limit As $x\to -\infty$, $$\frac{1}{x}\to 0,\quad \frac{1}{x^2}\to 0.$$ So the expression becomes $$\frac{0-0}{1+0+0}=0.$$ Therefore, $$\boxed{0}.$$ ### 💡 Pitfall guide Do not cancel the $x$ terms across a sum. For example, $(x-2)/(x^2+2x+1)$ is not the same as $1/(x+2+1/x)$. The safe method is to divide every term by the highest power in the denominator. Also, the negative sign in $x\to -\infty$ does not change the final answer here because the limit is zero. ### 🔄 Real-world variant If the denominator had the same degree as the numerator, the answer would depend on the leading coefficients. For example, $$\lim_{x\to-\infty}\frac{3x-2}{x^2+2x+1}=0,$$ but $$\lim_{x\to-\infty}\frac{3x-2}{x-1}=3.$$ If the numerator were $x^2-2$, then the limit would no longer be zero; the leading terms would control the result. ### 🔍 Related terms end behavior, rational function, degree comparison