\lim _{x\to 0}\frac{\sin \left(4x\right)\cdot \tan ^2\left(3x\right)\:+\:6x^2}{2x^2+\sin \left(3x\right)\cdot \cos \left(2x\right)} with only using trigonometry identities

To solve the limit \(\lim _{x\to 0}\frac{\sin \left(4x\right)\cdot \tan ^2\left(3x\right)\:+\:6x^2}{2x^2+\sin \left(3x\right)\cdot \cos \left(2x\right)}\) using trigonometric identities, let's analyze each component of the expression using small-angle approximations.

### Step 1: Small-Angle Approximations
For \(x\) close to 0, we can use the following approximations:

\[
\sin(y) \approx y \quad \text{and} \quad \tan(y) \approx y \quad \text{for small } y
\]

Substitute these approximations into the expression.

### Step 2: Simplify the Numerator
For the numerator \(\sin(4x) \cdot \tan^2(3x) + 6x^2\), we have:

\[
\sin(4x) \approx 4x
\]

\[
\tan^2(3x) \approx (3x)^2 = 9x^2
\]

Thus, the numerator simplifies to:

\[
\sin(4x) \cdot \tan^2(3x) + 6x^2 \approx 4x \cdot 9x^2 + 6x^2 = 36x^3 + 6x^2
\]

### Step 3: Simplify the Denominator
For the denominator \(2x^2 + \sin(3x) \cdot \cos(2x)\), we have:

\[
\sin(3x) \approx 3x
\]

\[
\cos(2x) \approx 1 \quad \text{(since \(\cos(y) \approx 1 - \frac{y^2}{2}\) and the higher order terms vanish as \(y \to 0\))}
\]

Thus, the denominator simplifies to:

\[
2x^2 + \sin(3x) \cdot \cos(2x) \approx 2x^2 + 3x \cdot 1 = 2x^2 + 3x
\]

### Step 4: Express

Learn how to evaluate the limit \(\lim _{x\to 0}\frac{\sin \left(4x\right)\cdot \tan ^2\left(3x\right)\:+\:6x^2}{2x^2+\sin \left(3x\right)\cdot \cos \left(2x\right)}\) using trigonometric identities and small-angle approximations. This problem involves analyzing trigonometric functions and applying small-angle approximations to simplify the expression.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation