Math Problem Statement

Calculate the limits of trigonometric expressions as x approaches 0.

Solution

The image contains a series of limit problems that involve trigonometric functions as x0x \to 0. Let’s solve these limits one by one.

1. limx0tan22x2x2\lim_{x \to 0} \frac{\tan^2 2x}{2x^2}

Using the approximation tanxx\tan x \approx x as x0x \to 0, we get: limx0tan2(2x)2x2=limx0(2x)22x2=4.\lim_{x \to 0} \frac{\tan^2(2x)}{2x^2} = \lim_{x \to 0} \frac{(2x)^2}{2x^2} = 4. So, the value is 4.

2. limx0sin6xxcos2x\lim_{x \to 0} \frac{\sin 6x}{x \cos 2x}

We use the approximations sinxx\sin x \approx x and cosx1\cos x \approx 1 as x0x \to 0: limx0sin(6x)xcos(2x)=limx06xx1=6.\lim_{x \to 0} \frac{\sin(6x)}{x \cos(2x)} = \lim_{x \to 0} \frac{6x}{x \cdot 1} = 6. So, the value is 6.

3. limx0sin2xtan8x\lim_{x \to 0} \frac{\sin 2x}{\tan 8x}

Using sinxx\sin x \approx x and tanxx\tan x \approx x as x0x \to 0: limx0sin(2x)tan(8x)=limx02x8x=28=14.\lim_{x \to 0} \frac{\sin(2x)}{\tan(8x)} = \lim_{x \to 0} \frac{2x}{8x} = \frac{2}{8} = \frac{1}{4}. So, the value is 14\frac{1}{4}.

4. limx0xsinx\lim_{x \to 0} \frac{x}{\sin x}

This is a standard limit that can be directly computed using the approximation sinxx\sin x \approx x as x0x \to 0: limx0xsinx=1.\lim_{x \to 0} \frac{x}{\sin x} = 1. So, the value is 1.

5. limx0tan3x5x\lim_{x \to 0} \frac{\tan 3x}{5x}

Using tanxx\tan x \approx x as x0x \to 0: limx0tan(3x)5x=limx03x5x=35.\lim_{x \to 0} \frac{\tan(3x)}{5x} = \lim_{x \to 0} \frac{3x}{5x} = \frac{3}{5}. So, the value is 35\frac{3}{5}.


Would you like any further details or explanations on these solutions?

Here are some related questions to consider:

  1. How do approximations like sinxx\sin x \approx x help in limit problems?
  2. What is the general approach to solving limits involving trigonometric functions?
  3. Can you explain why limx0xsinx=1\lim_{x \to 0} \frac{x}{\sin x} = 1 is important?
  4. What happens if the limits involve higher powers of xx or trigonometric terms?
  5. How would the limits change if xx were approaching values other than 0?

Tip: Always remember that trigonometric approximations sinxx\sin x \approx x and tanxx\tan x \approx x are very useful for limit problems as x0x \to 0.

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Math Problem Analysis

Mathematical Concepts

Limits
Trigonometric Functions
Calculus

Formulas

lim (x -> 0) (tan^2(2x) / 2x^2) = 4
lim (x -> 0) (sin(6x) / (x * cos(2x))) = 6
lim (x -> 0) (sin(2x) / tan(8x)) = 1/4
lim (x -> 0) (x / sin(x)) = 1
lim (x -> 0) (tan(3x) / 5x) = 3/5

Theorems

Squeeze Theorem
Small-angle approximations: sin(x) ≈ x, tan(x) ≈ x as x -> 0

Suitable Grade Level

Grades 11-12