Practice, Practice, … with solutions | Do More Math - Free Printable

Let’s go through each problem one by one, step by step. I’ll check the work shown and make sure everything is correct — because sometimes even if it looks right, a small mistake can change the answer.

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Problem 1:
> $\lim_{x \to -3} \frac{x+3}{x^2 - 4x - 21}$

First, factor the denominator:
$x^2 - 4x - 21 = (x - 7)(x + 3)$ → Yes, that’s correct.

So we have:
$\frac{x+3}{(x-7)(x+3)}$

Cancel $(x+3)$ from top and bottom → but only if $x \ne -3$, which is fine since we’re taking limit as x approaches -3, not at -3.

Left with: $\frac{1}{x - 7}$

Now plug in $x = -3$: $\frac{1}{-3 - 7} = \frac{1}{-10} = -\frac{1}{10}$

✔ The answer shown is correct: $-\frac{1}{10}$

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Problem 2:
> $\lim_{x \to \infty} \frac{x^2 + 500}{x^3 - 500}$

When x goes to infinity, look at highest powers: numerator has $x^2$, denominator has $x^3$. So this behaves like $\frac{x^2}{x^3} = \frac{1}{x}$, which goes to 0.

The solution says: $\lim_{x \to \infty} \frac{x^2}{x^3} = \lim_{x \to \infty} \frac{1}{x} = 0$

✔ Correct. Answer: 0

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Problem 3:
> $\lim_{x \to \infty} \frac{x^2 + 500}{x^2 + 1000}$

Highest power on top and bottom is $x^2$. Divide numerator and denominator by $x^2$:

→ $\frac{1 + \frac{500}{x^2}}{1 + \frac{1000}{x^2}}$

As $x \to \infty$, fractions go to 0 → so limit is $\frac{1}{1} = 1$

Solution says: $\lim_{x \to \infty} \frac{x^2}{x^2} = \lim_{x \to \infty} 1 = 1$

✔ Correct. Answer: 1

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Problem 4:
> $\lim_{x \to 0} \frac{x^2 + 500}{x^2 + 1000}$

This is NOT going to infinity — it’s going to 0! So just plug in x = 0:

Numerator: $0 + 500 = 500$

Denominator: $0 + 1000 = 1000$

So limit = $\frac{500}{1000} = \frac{1}{2}$

Solution says: $\lim_{x \to 0} \frac{500}{1000} = \frac{1}{2}$ → That’s correct, because when x=0, the $x^2$ terms vanish.

✔ Correct. Answer: $\frac{1}{2}$

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Problem 5:
> $\lim_{x \to \infty} \frac{x^4 - 5x + 200}{x^2 + 1000x + 1000}$

Highest power in numerator: $x^4$, denominator: $x^2$

So behaves like $\frac{x^4}{x^2} = x^2$, which goes to ∞ as x → ∞.

Solution says: $\lim_{x \to \infty} \frac{x^4}{x^2} = \lim_{x \to \infty} x^2 = \infty$

✔ Correct. Answer: ∞

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Problem 6:
> $\lim_{x \to 0} \frac{x^4 - 5x + 200}{x^2 + 1000x + 1000}$

Again, x → 0, so plug in x = 0:

Numerator: $0 - 0 + 200 = 200$

Denominator: $0 + 0 + 1000 = 1000$

Limit = $\frac{200}{1000} = \frac{1}{5}$

Solution says: $\lim_{x \to 0} \frac{200}{1000} = \frac{1}{5}$

✔ Correct. Answer: $\frac{1}{5}$

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All six problems are solved correctly in the worksheet!

Final Answer:
1. $-\frac{1}{10}$
2. $0$
3. $1$
4. $\frac{1}{2}$
5. $\infty$
6. $\frac{1}{5}$