Find the limit of 1 over x squared as x approaches negative infinity.

Step-by-step solution

1. 1

   Notice the even power erases the sign. For every $x\neq 0$ we have $x^2>0$, so it makes no difference whether $x$ runs off to $+\infty$ or $-\infty$: the denominator is positive either way. That is why this limit and the one at $+\infty$ must come out the same.

2. 2

   See that the denominator grows without bound. Since $x^2=|x|^2$ and $|x|\to\infty$ as $x\to-\infty$,

   $x^2\to+\infty$

3. 3

   Apply the fixed-over-unbounded rule. The numerator stays at $1$ while the denominator exceeds every bound, so the quotient is squeezed toward zero:

   $\frac{1}{x^2}\to 0$

   More precisely, for any $\varepsilon>0$ we have $1/x^2<\varepsilon$ as soon as $|x|>1/\sqrt{\varepsilon}$, which is the formal statement of the limit.

4. 4

   Check with numbers. At $x=-10$: $1/100=0.01$. At $x=-100$: $1/10000=0.0001$. At $x=-1000$: $10^{-6}$. The values shrink toward $0$ and stay positive throughout.

5. 5

   Record the answer and the direction of approach. The limit is $0$, approached from above — the function is never negative, so the graph flattens onto the $x$-axis from the top. The same argument gives $\lim_{x\to\pm\infty}x^{-n}=0$ for every $n>0$.