Math Problem Statement

Solution

Let's solve each problem from the image one by one.

1. Evaluate limx0(1sin2(θ))1/2\lim_{x \to 0} \left( 1 - \sin^2(\theta) \right)^{1/2}

This expression can be simplified using a trigonometric identity.

1sin2(θ)=cos2(θ)1 - \sin^2(\theta) = \cos^2(\theta)

So, the expression becomes:

limx0cos2(θ)=limx0cos(θ)\lim_{x \to 0} \sqrt{\cos^2(\theta)} = \lim_{x \to 0} |\cos(\theta)|

As θ\theta approaches 0, cos(0)=1\cos(0) = 1. Therefore, the limit is:

1\boxed{1}

2. Simplify the expression limx4x216x4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}

This expression involves a limit that can be simplified by factoring the numerator.

Factor the numerator x216x^2 - 16 as:

x216=(x4)(x+4)x^2 - 16 = (x - 4)(x + 4)

Thus, the expression becomes:

limx4(x4)(x+4)x4\lim_{x \to 4} \frac{(x - 4)(x + 4)}{x - 4}

Cancel out the common factor x4x - 4:

limx4(x+4)\lim_{x \to 4} (x + 4)

Now, substitute x=4x = 4:

4+4=84 + 4 = \boxed{8}

3. Evaluate the following limit limx1x21x2+3x4\lim_{x \to 1} \frac{x^2 - 1}{x^2 + 3x - 4}

First, factor both the numerator and the denominator:

  • Numerator: x21=(x1)(x+1)x^2 - 1 = (x - 1)(x + 1)
  • Denominator: x2+3x4=(x1)(x+4)x^2 + 3x - 4 = (x - 1)(x + 4)

The expression becomes:

limx1(x1)(x+1)(x1)(x+4)\lim_{x \to 1} \frac{(x - 1)(x + 1)}{(x - 1)(x + 4)}

Cancel out the common factor x1x - 1:

limx1x+1x+4\lim_{x \to 1} \frac{x + 1}{x + 4}

Substitute x=1x = 1:

1+11+4=25=25\frac{1 + 1}{1 + 4} = \frac{2}{5} = \boxed{\frac{2}{5}}

4. Evaluate the limit \lim_{x \to 4} \frac{(x - 4)/(x^2 - x - 12)}

First, factor the quadratic expression in the denominator:

  • Denominator: x2x12=(x4)(x+3)x^2 - x - 12 = (x - 4)(x + 3)

The expression becomes:

limx4x4(x4)(x+3)\lim_{x \to 4} \frac{x - 4}{(x - 4)(x + 3)}

Cancel out the common factor x4x - 4:

limx41x+3\lim_{x \to 4} \frac{1}{x + 3}

Substitute x=4x = 4:

14+3=17=17\frac{1}{4 + 3} = \frac{1}{7} = \boxed{\frac{1}{7}}

5. Evaluate the limit limxlnxx\lim_{x \to \infty} \frac{\ln x}{x}

For this limit, as xx \to \infty, lnx\ln x grows much slower than xx. Thus, the limit is:

limxlnxx=0=0\lim_{x \to \infty} \frac{\ln x}{x} = 0 = \boxed{0}

Would you like to continue with the remaining problems or have any other questions?

Here are 5 related questions:

  1. How would the limit in problem 1 change if θ\theta approaches π2\frac{\pi}{2}?
  2. What happens to the expression in problem 2 if we change the limit to x4x \to -4?
  3. How can L'Hôpital's rule be applied to problem 3?
  4. What does the expression in problem 4 evaluate to if the quadratic expression was not factorable?
  5. What would happen to the limit in problem 5 if we considered limx0+\lim_{x \to 0^+}?

Tip: Factoring and simplifying before taking limits often helps avoid indeterminate forms and simplifies the evaluation process.

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Math Problem Analysis

Mathematical Concepts

Limits
Trigonometric identities
Factoring
L'Hôpital's rule

Formulas

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Theorems

Limit properties

Suitable Grade Level

College level