Math Problem Statement

Compute the limits that exist, given that lim(x→0) f(x) = -1/2 and lim(x→0) g(x) = 1/2. Find the following limits: a) lim(x→0) [f(x) + g(x)] b) lim(x→0) [f(x) - 2g(x)] c) lim(x→0) [f(x) * g(x)] d) lim(x→0) [f(x) / g(x)]

Solution

Let's compute the given limits step by step. The problem provides us with the following information:

limx0f(x)=12andlimx0g(x)=12\lim_{x \to 0} f(x) = -\frac{1}{2} \quad \text{and} \quad \lim_{x \to 0} g(x) = \frac{1}{2}

We will apply limit properties to solve each part of the question.

a) limx0[f(x)+g(x)]\lim_{x \to 0} \left[f(x) + g(x)\right]

Using the sum rule for limits: limx0[f(x)+g(x)]=limx0f(x)+limx0g(x)\lim_{x \to 0} \left[f(x) + g(x)\right] = \lim_{x \to 0} f(x) + \lim_{x \to 0} g(x)

Substitute the given limits: limx0[f(x)+g(x)]=12+12=0\lim_{x \to 0} \left[f(x) + g(x)\right] = -\frac{1}{2} + \frac{1}{2} = 0

b) limx0[f(x)2g(x)]\lim_{x \to 0} \left[f(x) - 2g(x)\right]

Using the difference and scalar multiplication properties of limits: limx0[f(x)2g(x)]=limx0f(x)2limx0g(x)\lim_{x \to 0} \left[f(x) - 2g(x)\right] = \lim_{x \to 0} f(x) - 2\lim_{x \to 0} g(x)

Substitute the given limits: limx0[f(x)2g(x)]=122(12)=121=32\lim_{x \to 0} \left[f(x) - 2g(x)\right] = -\frac{1}{2} - 2\left(\frac{1}{2}\right) = -\frac{1}{2} - 1 = -\frac{3}{2}

c) limx0[f(x)×g(x)]\lim_{x \to 0} \left[f(x) \times g(x)\right]

Using the product rule for limits: limx0[f(x)×g(x)]=limx0f(x)×limx0g(x)\lim_{x \to 0} \left[f(x) \times g(x)\right] = \lim_{x \to 0} f(x) \times \lim_{x \to 0} g(x)

Substitute the given limits: limx0[f(x)×g(x)]=12×12=14\lim_{x \to 0} \left[f(x) \times g(x)\right] = -\frac{1}{2} \times \frac{1}{2} = -\frac{1}{4}

d) limx0f(x)g(x)\lim_{x \to 0} \frac{f(x)}{g(x)}

Using the quotient rule for limits, and noting that limx0g(x)0\lim_{x \to 0} g(x) \neq 0: limx0f(x)g(x)=limx0f(x)limx0g(x)\lim_{x \to 0} \frac{f(x)}{g(x)} = \frac{\lim_{x \to 0} f(x)}{\lim_{x \to 0} g(x)}

Substitute the given limits: limx0f(x)g(x)=1212=1\lim_{x \to 0} \frac{f(x)}{g(x)} = \frac{-\frac{1}{2}}{\frac{1}{2}} = -1

Summary of the limits:

  • a) limx0[f(x)+g(x)]=0\lim_{x \to 0} \left[f(x) + g(x)\right] = 0
  • b) limx0[f(x)2g(x)]=32\lim_{x \to 0} \left[f(x) - 2g(x)\right] = -\frac{3}{2}
  • c) limx0[f(x)×g(x)]=14\lim_{x \to 0} \left[f(x) \times g(x)\right] = -\frac{1}{4}
  • d) limx0f(x)g(x)=1\lim_{x \to 0} \frac{f(x)}{g(x)} = -1

Would you like any further clarification or more details on any of these steps?

Here are 5 related questions for practice:

  1. What is the limit of f(x)+3g(x)f(x) + 3g(x) as x0x \to 0?
  2. Compute limx0[2f(x)3g(x)]\lim_{x \to 0} \left[2f(x) - 3g(x)\right] using the same given limits.
  3. What is the result of limx0[f(x)×g(x)g(x)]\lim_{x \to 0} \left[\frac{f(x) \times g(x)}{g(x)}\right]?
  4. How would you find limx0[f(x)2g(x)2]\lim_{x \to 0} \left[f(x)^2 - g(x)^2\right]?
  5. Compute limx0[3f(x)g(x)]\lim_{x \to 0} \left[3f(x) \cdot g(x)\right] using the limits provided.

Tip: Remember to use limit laws like the sum, difference, product, and quotient rules for simplifying complex expressions.

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Math Problem Analysis

Mathematical Concepts

Limits
Functions
Calculus

Formulas

lim(x→0) [f(x) + g(x)] = lim(x→0) f(x) + lim(x→0) g(x)
lim(x→0) [f(x) - 2g(x)] = lim(x→0) f(x) - 2*lim(x→0) g(x)
lim(x→0) [f(x) * g(x)] = lim(x→0) f(x) * lim(x→0) g(x)
lim(x→0) [f(x) / g(x)] = lim(x→0) f(x) / lim(x→0) g(x) when lim(x→0) g(x) ≠ 0

Theorems

Limit Laws (Sum, Product, Quotient, and Constant Multiple Rules)

Suitable Grade Level

Undergraduate Calculus or Advanced High School Mathematics