Evaluate $\lim_{t \rightarrow 0} \frac{t^3}{\tan^3(2t)}$.

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Introduction

In mathematics, the concept of limits is a fundamental aspect of calculus, and it plays a crucial role in understanding various mathematical functions and their behavior. The limit of a function as the input or independent variable approaches a specific value is a measure of the function's behavior at that point. In this article, we will evaluate the limit of the function $\frac{t^3}{\tan^3(2t)}$ as $t$ approaches 0.

Background

The function $\tan^3(2t)$ is a trigonometric function, and it is known to be undefined at certain points. In this case, we are dealing with the limit as $t$ approaches 0, which is a critical point for the function. The function $\tan^3(2t)$ can be rewritten as $(\tan(2t))^3$, which is a composite function. The inner function $\tan(2t)$ is a trigonometric function, and the outer function is a power function.

The Limit

To evaluate the limit of the function $\frac{t^3}{\tan^3(2t)}$ as $t$ approaches 0, we can use the following approach:

$\lim_{t \rightarrow 0} \frac{t^3}{\tan^3(2t)} = \lim_{t \rightarrow 0} \frac{t^3}{((\tan(2t))^3)}$

We can rewrite the function as:

$\lim_{t \rightarrow 0} \frac{t^3}{((\tan(2t))^3)} = \lim_{t \rightarrow 0} \frac{t^3}{((\tan(2t))^3)} \cdot \frac{1}{\tan(2t)} \cdot \frac{\tan(2t)}{2t} \cdot \frac{2t}{t}$

This can be simplified as:

$\lim_{t \rightarrow 0} \frac{t^3}{((\tan(2t))^3)} = \lim_{t \rightarrow 0} \frac{t^3}{((\tan(2t))^3)} \cdot \frac{1}{\tan(2t)} \cdot \frac{\tan(2t)}{2t} \cdot \frac{2}{1}$

Simplifying the Expression

We can simplify the expression further by using the following trigonometric identity:

$\tan(2t) = \frac{2\tan(t)}{1 - \tan^2(t)}$

Substituting this identity into the expression, we get:

$\lim_{t \rightarrow 0} \frac{t^3}{((\tan(2t))^3)} = \lim_{t \rightarrow 0} \frac{t^3}{((\tan(2t))^3)} \cdot \frac{1}{\tan(2t)} \cdot \frac{\tan(2t)}{2t} \cdot \frac{2}{1} \cdot \frac{1 - \tan^2(t)}{2\tan(t)}$

Evaluating the Limit

We can evaluate the limit by substituting $t = 0$ into the expression:

$\lim_{t \rightarrow 0} \frac{t^3}{((\tan(2t))^3)} = \frac{0^3}{((\tan(0))^3)} \cdot \frac{1}{\tan(0)} \cdot \frac{\tan(0)}{2(0)} \cdot \frac{2}{1} \cdot \frac{1 - \tan^2(0)}{2\tan(0)}$

Since $\tan(0) = 0$, we can simplify the expression further:

$\lim_{t \rightarrow 0} \frac{t^3}{((\tan(2t))^3)} = \frac{0}{0} \cdot \frac{1}{0} \cdot \frac{0}{0} \cdot \frac{2}{1} \cdot \frac{1 - 0^2}{2(0)}$

Applying L'Hopital's Rule

Since we have an indeterminate form of $\frac{0}{0}$, we can apply L'Hopital's rule to evaluate the limit:

$\lim_{t \rightarrow 0} \frac{t^3}{((\tan(2t))^3)} = \lim_{t \rightarrow 0} \frac{3t^2}{3(\tan(2t))^2 \cdot \sec^2(2t)}$

Evaluating the Limit Again

We can evaluate the limit again by substituting $t = 0$ into the expression:

$\lim_{t \rightarrow 0} \frac{t^3}{((\tan(2t))^3)} = \frac{3(0)^2}{3((\tan(0))^2) \cdot \sec^2(0)}$

Since $\tan(0) = 0$ and $\sec(0) = 1$, we can simplify the expression further:

$\lim_{t \rightarrow 0} \frac{t^3}{((\tan(2t))^3)} = \frac{0}{0 \cdot 1}$

Applying L'Hopital's Rule Again

Since we have an indeterminate form of $\frac{0}{0}$ again, we can apply L'Hopital's rule again to evaluate the limit:

$\lim_{t \rightarrow 0} \frac{t^3}{((\tan(2t))^3)} = \lim_{t \rightarrow 0} \frac{6t}{6(\tan(2t)) \cdot \sec^2(2t) \cdot \tan(2t) \cdot \sec^2(2t)}$

Evaluating the Limit Again

We can evaluate the limit again by substituting $t = 0$ into the expression:

$\lim_{t \rightarrow 0} \frac{t^3}{((\tan(2t))^3)} = \frac{6(0)}{6((\tan(0))) \cdot \sec^2(0) \cdot \tan(0) \cdot \sec^2(0)}$

Since $\tan(0) = 0$ and $\sec(0) = 1$, we can simplify the expression further:

$\lim_{t \rightarrow 0} \frac{t^3}{((\tan(2t))^3)} = \frac{0}{0 \cdot 1 \cdot 0 \cdot 1}$

Conclusion

We have evaluated the limit of the function $\frac{t^3}{\tan^3(2t)}$ as $t$ approaches 0 using L'Hopital's rule. The final answer is $\boxed{0}$.

Final Answer

The final answer is $\boxed{0}$.

References

• [1] L'Hopital, G. F. A. (1696). Analyse des Infiniment Petits pour l'Intelligence des Lignes Courbes.
• [2] Taylor, B. (1715). Methodus Incrementorum Directa et Inversa.
• [3] Maclaurin, C. (1742). A Treatise of Algebra.

Keywords

• Limit
• L'Hopital's rule
• Trigonometric functions
• Indeterminate form
• Calculus

Categories

• Mathematics
• Calculus
• Limits
• L'Hopital's rule
• Trigonometric functions
  

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Introduction

In our previous article, we evaluated the limit of the function $\frac{t^3}{\tan^3(2t)}$ as $t$ approaches 0 using L'Hopital's rule. In this article, we will answer some frequently asked questions related to this topic.

Q: What is L'Hopital's rule?

A: L'Hopital's rule is a mathematical technique used to evaluate the limit of a function as the input or independent variable approaches a specific value. It is used to solve indeterminate forms of the type $\frac{0}{0}$ or $\frac{\infty}{\infty}$.

Q: Why do we need L'Hopital's rule?

A: We need L'Hopital's rule to evaluate the limit of a function when the function is in an indeterminate form. This is because the function is not defined at the point where the limit is being evaluated.

Q: How do we apply L'Hopital's rule?

A: To apply L'Hopital's rule, we need to take the derivative of the numerator and the derivative of the denominator separately. We then evaluate the limit of the ratio of the derivatives.

Q: What is the significance of the limit of $\frac{t^3}{\tan^3(2t)}$?

A: The limit of $\frac{t^3}{\tan^3(2t)}$ is significant because it helps us understand the behavior of the function as $t$ approaches 0. This is important in calculus and other areas of mathematics.

Q: Can we use L'Hopital's rule for other types of limits?

A: Yes, we can use L'Hopital's rule for other types of limits, such as limits involving exponential functions, logarithmic functions, and trigonometric functions.

Q: What are some common mistakes to avoid when using L'Hopital's rule?

A: Some common mistakes to avoid when using L'Hopital's rule include:

• Not taking the derivative of the numerator and the denominator separately
• Not evaluating the limit of the ratio of the derivatives
• Not checking if the limit is still in an indeterminate form after applying L'Hopital's rule

Q: How do we check if the limit is still in an indeterminate form after applying L'Hopital's rule?

A: To check if the limit is still in an indeterminate form after applying L'Hopital's rule, we need to evaluate the limit of the ratio of the derivatives. If the limit is still in an indeterminate form, we need to apply L'Hopital's rule again.

Q: Can we use L'Hopital's rule for limits involving complex functions?

A: Yes, we can use L'Hopital's rule for limits involving complex functions. However, we need to be careful when dealing with complex functions, as they can have different properties than real functions.

Q: What are some real-world applications of L'Hopital's rule?

A: L'Hopital's rule has many real-world applications, including:

• Physics: L'Hopital's rule is used to evaluate the limit of functions that describe physical phenomena, such as the motion of objects.
• Engineering: L'Hopital's rule is used to evaluate the limit of functions that describe the behavior of systems, such as electrical circuits.
• Economics: L'Hopital's rule is used to evaluate the limit of functions that describe economic phenomena, such as the behavior of markets.

Conclusion

In this article, we have answered some frequently asked questions related to the limit of $\frac{t^3}{\tan^3(2t)}$. We have also discussed the significance of L'Hopital's rule and its applications in various fields.

Final Answer

The final answer is $\boxed{0}$.

References

• [1] L'Hopital, G. F. A. (1696). Analyse des Infiniment Petits pour l'Intelligence des Lignes Courbes.
• [2] Taylor, B. (1715). Methodus Incrementorum Directa et Inversa.
• [3] Maclaurin, C. (1742). A Treatise of Algebra.

Keywords

• Limit
• L'Hopital's rule
• Trigonometric functions
• Indeterminate form
• Calculus

Categories

• Mathematics
• Calculus
• Limits
• L'Hopital's rule
• Trigonometric functions