Let’s solve each limit problem one by one with clear explanations.
---
1) Evaluate
\[
\lim_{x \to 2} \frac{x^2 + 5x + 6}{x + 2}
\]
Step 1: Direct substitution
Plug in \( x = 2 \):
Numerator: \( 2^2 + 5(2) + 6 = 4 + 10 + 6 = 20 \)
Denominator: \( 2 + 2 = 4 \)
So, \( \frac{20}{4} = 5 \)
✔ Since the function is defined and continuous at \( x = 2 \), we can directly substitute.
Answer:
\[
\boxed{5}
\]
---
2) Evaluate
\[
\lim_{x \to 2} \frac{x^3 - 8}{x - 2}
\]
Step 1: Recognize the form
This is a classic “difference of cubes” indeterminate form \( \frac{0}{0} \) when \( x = 2 \).
Recall:
\[
a^3 - b^3 = (a - b)(a^2 + ab + b^2)
\]
Here, \( x^3 - 8 = x^3 - 2^3 = (x - 2)(x^2 + 2x + 4) \)
Step 2: Factor and simplify
\[
\frac{x^3 - 8}{x - 2} = \frac{(x - 2)(x^2 + 2x + 4)}{x - 2} = x^2 + 2x + 4 \quad \text{(for } x \ne 2\text{)}
\]
Step 3: Take the limit
Now plug in \( x = 2 \):
\[
2^2 + 2(2) + 4 = 4 + 4 + 4 = 12
\]
✔ The discontinuity at \( x=2 \) is removable; after simplifying, the limit exists.
Answer:
\[
\boxed{12}
\]
---
3) Evaluate
\[
\lim_{x \to \infty} \frac{x^2 + 1}{1 + 2x^3}
\]
Step 1: Compare degrees
Numerator degree: 2
Denominator degree: 3
Since denominator has higher degree, limit → 0 as \( x \to \infty \)
Step 2: Divide numerator and denominator by highest power of \( x \) in denominator — which is \( x^3 \)
\[
\frac{x^2 + 1}{1 + 2x^3} = \frac{\frac{x^2}{x^3} + \frac{1}{x^3}}{\frac{1}{x^3} + 2} = \frac{\frac{1}{x} + \frac{1}{x^3}}{\frac{1}{x^3} + 2}
\]
As \( x \to \infty \), all terms with \( x \) in denominator → 0:
\[
\frac{0 + 0}{0 + 2} = \frac{0}{2} = 0
\]
Answer:
\[
\boxed{0}
\]
---
4) Evaluate
\[
\lim_{x \to \infty} \frac{4x^2 - 3x}{5x^2 - 2}
\]
Step 1: Degrees are equal (both 2) → limit is ratio of leading coefficients.
Leading coefficient of numerator: 4
Leading coefficient of denominator: 5
So, limit = \( \frac{4}{5} \)
Step 2: Verify by dividing numerator and denominator by \( x^2 \)
\[
\frac{4x^2 - 3x}{5x^2 - 2} = \frac{4 - \frac{3}{x}}{5 - \frac{2}{x^2}}
\]
As \( x \to \infty \), \( \frac{3}{x} \to 0 \), \( \frac{2}{x^2} \to 0 \):
\[
\frac{4 - 0}{5 - 0} = \frac{4}{5}
\]
Answer:
\[
\boxed{\frac{4}{5}}
\]
---
##
✔ Final Answers:
1) \( \boxed{5} \)
2) \( \boxed{12} \)
3) \( \boxed{0} \)
4) \( \boxed{\frac{4}{5}} \)
Parent Tip: Review the logic above to help your child master the concept of simple limits worksheet.
