Calculus worksheet - limits and functions - Free Printable

Let's solve each of these limit problems step by step.

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1) Evaluate $\lim_{x \to 2} \frac{x^2 + 5x + 6}{x + 2}$

Step 1: Try direct substitution.

Plug in $x = 2$:

$$
\frac{(2)^2 + 5(2) + 6}{2 + 2} = \frac{4 + 10 + 6}{4} = \frac{20}{4} = 5
$$

✔ The function is defined at $x = 2$, so the limit exists and equals the value of the function.

Answer: $\boxed{5}$

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2) Evaluate $\lim_{x \to 2} \frac{x^3 - 8}{x - 2}$

Step 1: Notice that $x^3 - 8$ is a difference of cubes:
$$
x^3 - 8 = (x - 2)(x^2 + 2x + 4)
$$

So we can factor:

$$
\frac{x^3 - 8}{x - 2} = \frac{(x - 2)(x^2 + 2x + 4)}{x - 2}
$$

For $x \ne 2$, we can cancel $x - 2$:

$$
= x^2 + 2x + 4
$$

Now take the limit as $x \to 2$:

$$
\lim_{x \to 2} (x^2 + 2x + 4) = (2)^2 + 2(2) + 4 = 4 + 4 + 4 = 12
$$

Answer: $\boxed{12}$

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3) Evaluate $\lim_{x \to \infty} \frac{x^2 + 1}{1 + 2x^3}$

Step 1: This is an infinite limit. Compare degrees of numerator and denominator.

- Numerator: degree 2
- Denominator: degree 3

Since the degree of the denominator is higher, the limit will go to 0.

But let’s confirm by dividing numerator and denominator by the highest power of $x$ in the denominator, which is $x^3$:

$$
\frac{x^2 + 1}{1 + 2x^3} = \frac{\frac{x^2}{x^3} + \frac{1}{x^3}}{\frac{1}{x^3} + \frac{2x^3}{x^3}} = \frac{\frac{1}{x} + \frac{1}{x^3}}{\frac{1}{x^3} + 2}
$$

As $x \to \infty$, all terms with $x$ in the denominator go to 0:

$$
\frac{0 + 0}{0 + 2} = \frac{0}{2} = 0
$$

Answer: $\boxed{0}$

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4) Evaluate $\lim_{x \to \infty} \frac{4x^2 - 3x}{5x^2 - 2}$

Step 1: Both numerator and denominator are degree 2 polynomials.

When degrees are equal, the limit is the ratio of leading coefficients.

Leading term in numerator: $4x^2$

Leading term in denominator: $5x^2$

So:

$$
\lim_{x \to \infty} \frac{4x^2 - 3x}{5x^2 - 2} = \frac{4}{5}
$$

To verify, divide numerator and denominator by $x^2$:

$$
\frac{4 - \frac{3}{x}}{5 - \frac{2}{x^2}} \to \frac{4 - 0}{5 - 0} = \frac{4}{5}
$$

Answer: $\boxed{\frac{4}{5}}$

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✔ Final Answers:

1) $\boxed{5}$
2) $\boxed{12}$
3) $\boxed{0}$
4) $\boxed{\frac{4}{5}}$