Evaluate The Limit:${ \lim _{x \rightarrow \pi^{+}}\left(\tan \frac{x}{2}\right) }$

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Introduction

In mathematics, evaluating limits is a crucial concept that helps us understand the behavior of functions as the input values approach a specific point. The given limit, $\lim _{x \rightarrow \pi^{+}}\left(\tan \frac{x}{2}\right)$, involves the tangent function and a specific value of $x$ approaching $\pi$ from the right. In this article, we will delve into the world of trigonometric functions, explore the properties of the tangent function, and evaluate the given limit.

Understanding the Tangent Function

The tangent function, denoted by $\tan x$, is a fundamental trigonometric function that is defined as the ratio of the sine and cosine functions. Mathematically, $\tan x = \frac{\sin x}{\cos x}$. The tangent function has a periodic nature, with a period of $\pi$, and its range is all real numbers.

Properties of the Tangent Function

The tangent function has several important properties that are essential in evaluating limits. One of the key properties is that the tangent function is continuous everywhere, except at odd multiples of $\frac{\pi}{2}$. This means that the tangent function can be evaluated at any point, except at these specific values.

Evaluating the Limit

To evaluate the given limit, $\lim _{x \rightarrow \pi^{+}}\left(\tan \frac{x}{2}\right)$, we need to consider the behavior of the tangent function as $x$ approaches $\pi$ from the right. Since the tangent function is continuous everywhere, except at odd multiples of $\frac{\pi}{2}$, we can evaluate the limit by substituting $x = \pi$ into the expression.

Substitution Method

Using the substitution method, we can evaluate the limit as follows:

$$\lim _{x \rightarrow \pi^{+}}\left(\tan \frac{x}{2}\right) = \tan \frac{\pi}{2}$$

However, this expression is undefined, as the tangent function is not defined at $\frac{\pi}{2}$. Therefore, we need to use a different approach to evaluate the limit.

L'Hopital's Rule

L'Hopital's rule is a powerful tool for evaluating limits that involve indeterminate forms. In this case, we can apply L'Hopital's rule to evaluate the limit as follows:

$$\lim _{x \rightarrow \pi^{+}}\left(\tan \frac{x}{2}\right) = \lim _{x \rightarrow \pi^{+}}\left(\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}\right)$$

Using L'Hopital's rule, we can rewrite the expression as:

$$\lim _{x \rightarrow \pi^{+}}\left(\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}\right) = \lim _{x \rightarrow \pi^{+}}\left(\frac{\frac{1}{2}\cos \frac{x}{2}}{-\frac{1}{2}\sin \frac{x}{2}}\right)$$

Simplifying the expression, we get:

$$\lim _{x \rightarrow \pi^{+}}\left(\frac{\frac{1}{2}\cos \frac{x}{2}}{-\frac{1}{2}\sin \frac{x}{2}}\right) = \lim _{x \rightarrow \pi^{+}}\left(-\cot \frac{x}{2}\right)$$

Evaluating the Limit Using the Cotangent Function

The cotangent function, denoted by $\cot x$, is the reciprocal of the tangent function. Mathematically, $\cot x = \frac{1}{\tan x}$. Using the cotangent function, we can rewrite the expression as:

$$\lim _{x \rightarrow \pi^{+}}\left(-\cot \frac{x}{2}\right) = -\cot \frac{\pi}{2}$$

Since the cotangent function is continuous everywhere, except at odd multiples of $\frac{\pi}{2}$, we can evaluate the limit by substituting $x = \pi$ into the expression.

Substitution Method

Using the substitution method, we can evaluate the limit as follows:

$$-\cot \frac{\pi}{2} = -\cot \left(\frac{\pi}{2}\right)$$

Since the cotangent function is not defined at $\frac{\pi}{2}$, we need to use a different approach to evaluate the limit.

L'Hopital's Rule

L'Hopital's rule is a powerful tool for evaluating limits that involve indeterminate forms. In this case, we can apply L'Hopital's rule to evaluate the limit as follows:

$$-\cot \frac{\pi}{2} = -\lim _{x \rightarrow \frac{\pi}{2}^{-}}\left(\tan x\right)$$

Using L'Hopital's rule, we can rewrite the expression as:

$$-\lim _{x \rightarrow \frac{\pi}{2}^{-}}\left(\tan x\right) = -\lim _{x \rightarrow \frac{\pi}{2}^{-}}\left(\frac{\sin x}{\cos x}\right)$$

Simplifying the expression, we get:

$$-\lim _{x \rightarrow \frac{\pi}{2}^{-}}\left(\frac{\sin x}{\cos x}\right) = -\lim _{x \rightarrow \frac{\pi}{2}^{-}}\left(\frac{\cos x}{-\sin x}\right)$$

Evaluating the Limit Using the Sine and Cosine Functions

The sine and cosine functions are fundamental trigonometric functions that are defined as the ratios of the opposite and adjacent sides of a right triangle. Mathematically, $\sin x = \frac{\text{opposite}}{\text{hypotenuse}}$ and $\cos x = \frac{\text{adjacent}}{\text{hypotenuse}}$. Using the sine and cosine functions, we can rewrite the expression as:

$$-\lim _{x \rightarrow \frac{\pi}{2}^{-}}\left(\frac{\cos x}{-\sin x}\right) = -\lim _{x \rightarrow \frac{\pi}{2}^{-}}\left(\frac{\frac{\text{adjacent}}{\text{hypotenuse}}}{-\frac{\text{opposite}}{\text{hypotenuse}}}\right)$$

Simplifying the expression, we get:

$$-\lim _{x \rightarrow \frac{\pi}{2}^{-}}\left(\frac{\frac{\text{adjacent}}{\text{hypotenuse}}}{-\frac{\text{opposite}}{\text{hypotenuse}}}\right) = -\lim _{x \rightarrow \frac{\pi}{2}^{-}}\left(\frac{\text{adjacent}}{-\text{opposite}}\right)$$

Evaluating the Limit Using the Right Triangle

Using the right triangle, we can evaluate the limit as follows:

$$-\lim _{x \rightarrow \frac{\pi}{2}^{-}}\left(\frac{\text{adjacent}}{-\text{opposite}}\right) = -\lim _{x \rightarrow \frac{\pi}{2}^{-}}\left(\frac{\text{adjacent}}{-\text{opposite}}\right)$$

Since the adjacent side approaches 0 and the opposite side approaches 1 as $x$ approaches $\frac{\pi}{2}$ from the left, we can evaluate the limit as follows:

$$-\lim _{x \rightarrow \frac{\pi}{2}^{-}}\left(\frac{\text{adjacent}}{-\text{opposite}}\right) = -\frac{0}{-1}$$

Simplifying the expression, we get:

$$-\frac{0}{-1} = 0$$

Conclusion

In conclusion, we have evaluated the limit $\lim _{x \rightarrow \pi^{+}}\left(\tan \frac{x}{2}\right)$ using various methods, including the substitution method, L'Hopital's rule, and the sine and cosine functions. We have shown that the limit approaches 0 as $x$ approaches $\pi$ from the right. This result is consistent with the properties of the tangent function and the behavior of the sine and cosine functions as $x$ approaches $\pi$ from the right.

Final Answer

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

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Introduction

In our previous article, we evaluated the limit $\lim _{x \rightarrow \pi^{+}}\left(\tan \frac{x}{2}\right)$ using various methods, including the substitution method, L'Hopital's rule, and the sine and cosine functions. In this article, we will answer some frequently asked questions (FAQs) related to this limit.

Q: What is the significance of the limit $\lim _{x \rightarrow \pi^{+}}\left(\tan \frac{x}{2}\right)$?

A: The limit $\lim _{x \rightarrow \pi^{+}}\left(\tan \frac{x}{2}\right)$ is significant because it helps us understand the behavior of the tangent function as $x$ approaches $\pi$ from the right. This limit is also related to the properties of the tangent function and the behavior of the sine and cosine functions as $x$ approaches $\pi$ from the right.

Q: Why is the limit $\lim _{x \rightarrow \pi^{+}}\left(\tan \frac{x}{2}\right)$ important in mathematics?

A: The limit $\lim _{x \rightarrow \pi^{+}}\left(\tan \frac{x}{2}\right)$ is important in mathematics because it helps us understand the behavior of trigonometric functions, which are essential in various mathematical disciplines, such as calculus, algebra, and geometry.

Q: Can you explain the concept of L'Hopital's rule in more detail?

A: L'Hopital's rule is a powerful tool for evaluating limits that involve indeterminate forms. It states that if a limit is in the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, we can rewrite the expression as a ratio of derivatives and evaluate the limit of the ratio.

Q: How do you apply L'Hopital's rule to evaluate the limit $\lim _{x \rightarrow \pi^{+}}\left(\tan \frac{x}{2}\right)$?

A: To apply L'Hopital's rule to evaluate the limit $\lim _{x \rightarrow \pi^{+}}\left(\tan \frac{x}{2}\right)$, we need to rewrite the expression as a ratio of derivatives and evaluate the limit of the ratio. We can do this by using the chain rule and the product rule to find the derivatives of the numerator and denominator.

Q: What is the relationship between the tangent function and the sine and cosine functions?

A: The tangent function is related to the sine and cosine functions through the following identity: $\tan x = \frac{\sin x}{\cos x}$. This identity is essential in evaluating limits that involve the tangent function.

Q: Can you explain the concept of the cotangent function in more detail?

A: The cotangent function is the reciprocal of the tangent function. It is defined as $\cot x = \frac{1}{\tan x}$. The cotangent function is also related to the sine and cosine functions through the following identity: $\cot x = \frac{\cos x}{\sin x}$.

Q: How do you evaluate the limit $\lim _{x \rightarrow \pi^{+}}\left(\cot \frac{x}{2}\right)$?

A: To evaluate the limit $\lim _{x \rightarrow \pi^{+}}\left(\cot \frac{x}{2}\right)$, we can use the identity $\cot x = \frac{\cos x}{\sin x}$ and rewrite the expression as $\frac{\cos \frac{x}{2}}{\sin \frac{x}{2}}$. We can then evaluate the limit by using L'Hopital's rule or by using the properties of the sine and cosine functions.

Q: What is the final answer to the limit $\lim _{x \rightarrow \pi^{+}}\left(\tan \frac{x}{2}\right)$?

A: The final answer to the limit $\lim _{x \rightarrow \pi^{+}}\left(\tan \frac{x}{2}\right)$ is $\boxed{0}$.

Conclusion

In conclusion, we have answered some frequently asked questions related to the limit $\lim _{x \rightarrow \pi^{+}}\left(\tan \frac{x}{2}\right)$. We have explained the significance of this limit, the importance of L'Hopital's rule, and the relationship between the tangent function and the sine and cosine functions. We have also provided step-by-step solutions to evaluate the limit using various methods.