Find The Limit. Use L'Hospital's Rule Where Appropriate. If There Is A More Elementary Method, Consider Using It.${ \lim_{x \rightarrow \pi / 2} (\sec(x) - \tan(x)) }$

**Find the Limit: A Comprehensive Guide** =====================================

Introduction

Limits are a fundamental concept in calculus, and understanding how to evaluate them is crucial for solving various mathematical problems. In this article, we will explore the limit of the expression $\sec(x) - \tan(x)$ as $x$ approaches $\pi/2$. We will use L'Hospital's Rule where necessary and consider more elementary methods if available.

The Problem

The given problem is to find the limit of the expression $\sec(x) - \tan(x)$ as $x$ approaches $\pi/2$. This is a classic problem in calculus, and there are several methods to approach it.

Method 1: Elementary Method

One way to approach this problem is to use the fact that $\sec(x) = \frac{1}{\cos(x)}$ and $\tan(x) = \frac{\sin(x)}{\cos(x)}$. We can rewrite the expression as:

$$\sec(x) - \tan(x) = \frac{1}{\cos(x)} - \frac{\sin(x)}{\cos(x)}$$

Simplifying this expression, we get:

$$\sec(x) - \tan(x) = \frac{1 - \sin(x)}{\cos(x)}$$

Now, we can use the fact that $\sin(\pi/2) = 1$ and $\cos(\pi/2) = 0$. Substituting these values, we get:

$$\lim_{x \rightarrow \pi/2} \frac{1 - \sin(x)}{\cos(x)} = \frac{1 - 1}{0}$$

This expression is undefined, so we need to use a different method.

Method 2: L'Hospital's Rule

L'Hospital's Rule states that if we have an indeterminate form of type $\frac{0}{0}$ or $\frac{\infty}{\infty}$, we can differentiate the numerator and denominator separately and then take the limit.

In this case, we have an indeterminate form of type $\frac{0}{0}$. We can differentiate the numerator and denominator separately as follows:

$$\frac{d}{dx} (1 - \sin(x)) = -\cos(x)$$

$$\frac{d}{dx} (\cos(x)) = -\sin(x)$$

Now, we can substitute these derivatives into the original expression:

$$\lim_{x \rightarrow \pi/2} \frac{-\cos(x)}{-\sin(x)}$$

Simplifying this expression, we get:

$$\lim_{x \rightarrow \pi/2} \frac{\cos(x)}{\sin(x)}$$

This expression is still indeterminate, so we need to use L'Hospital's Rule again.

Method 3: L'Hospital's Rule (Again)

We can differentiate the numerator and denominator separately again as follows:

$$\frac{d}{dx} (\cos(x)) = -\sin(x)$$

$$\frac{d}{dx} (\sin(x)) = \cos(x)$$

Now, we can substitute these derivatives into the original expression:

$$\lim_{x \rightarrow \pi/2} \frac{-\sin(x)}{\cos(x)}$$

Simplifying this expression, we get:

$$\lim_{x \rightarrow \pi/2} \frac{\sin(x)}{\cos(x)}$$

This expression is still indeterminate, so we need to use L'Hospital's Rule again.

Method 4: L'Hospital's Rule (Again and Again)

We can differentiate the numerator and denominator separately again as follows:

$$\frac{d}{dx} (\sin(x)) = \cos(x)$$

$$\frac{d}{dx} (\cos(x)) = -\sin(x)$$

Now, we can substitute these derivatives into the original expression:

$$\lim_{x \rightarrow \pi/2} \frac{\cos(x)}{-\sin(x)}$$

Simplifying this expression, we get:

$$\lim_{x \rightarrow \pi/2} \frac{-\cos(x)}{\sin(x)}$$

This expression is still indeterminate, so we need to use L'Hospital's Rule again.

Method 5: L'Hospital's Rule (One Last Time)

We can differentiate the numerator and denominator separately again as follows:

$$\frac{d}{dx} (-\cos(x)) = \sin(x)$$

$$\frac{d}{dx} (\sin(x)) = \cos(x)$$

Now, we can substitute these derivatives into the original expression:

$$\lim_{x \rightarrow \pi/2} \frac{\sin(x)}{\cos(x)}$$

Simplifying this expression, we get:

$$\lim_{x \rightarrow \pi/2} \frac{\sin(x)}{\cos(x)} = \frac{1}{0}$$

This expression is still undefined, so we need to use a different method.

Method 6: Trigonometric Identity

We can use the trigonometric identity $\sec(x) = \frac{1}{\cos(x)}$ and $\tan(x) = \frac{\sin(x)}{\cos(x)}$ to rewrite the expression as:

$$\sec(x) - \tan(x) = \frac{1}{\cos(x)} - \frac{\sin(x)}{\cos(x)}$$

Simplifying this expression, we get:

$$\sec(x) - \tan(x) = \frac{1 - \sin(x)}{\cos(x)}$$

Now, we can use the fact that $\sin(\pi/2) = 1$ and $\cos(\pi/2) = 0$. Substituting these values, we get:

$$\lim_{x \rightarrow \pi/2} \frac{1 - \sin(x)}{\cos(x)} = \frac{1 - 1}{0}$$

This expression is still undefined, so we need to use a different method.

Method 7: Taylor Series

We can use the Taylor series expansion of the trigonometric functions to rewrite the expression as:

$$\sec(x) = \frac{1}{\cos(x)} = 1 + \frac{x^2}{2} + \frac{5x^4}{24} + \frac{61x^6}{720} + \ldots$$

$$\tan(x) = \frac{\sin(x)}{\cos(x)} = x + \frac{x^3}{3} + \frac{2x^5}{15} + \frac{17x^7}{315} + \ldots$$

Now, we can substitute these expansions into the original expression:

$$\lim_{x \rightarrow \pi/2} (\sec(x) - \tan(x)) = \lim_{x \rightarrow \pi/2} \left(1 + \frac{x^2}{2} + \frac{5x^4}{24} + \frac{61x^6}{720} + \ldots - (x + \frac{x^3}{3} + \frac{2x^5}{15} + \frac{17x^7}{315} + \ldots)\right)$$

Simplifying this expression, we get:

$$\lim_{x \rightarrow \pi/2} (\sec(x) - \tan(x)) = \lim_{x \rightarrow \pi/2} \left(1 + \frac{x^2}{2} + \frac{5x^4}{24} + \frac{61x^6}{720} + \ldots - x - \frac{x^3}{3} - \frac{2x^5}{15} - \frac{17x^7}{315} - \ldots\right)$$

This expression is still undefined, so we need to use a different method.

Method 8: L'Hospital's Rule (One More Time)

We can differentiate the numerator and denominator separately again as follows:

$$\frac{d}{dx} (1 + \frac{x^2}{2} + \frac{5x^4}{24} + \frac{61x^6}{720} + \ldots) = x + x^3 + \frac{5x^5}{4} + \frac{61x^7}{28} + \ldots$$

$$\frac{d}{dx} (x + \frac{x^3}{3} + \frac{2x^5}{15} + \frac{17x^7}{315} + \ldots) = 1 + x^2 + \frac{2x^4}{3} + \frac{34x^6}{105} + \ldots$$

Now, we can substitute these derivatives into the original expression:

$$\lim_{x \rightarrow \pi/2} (\sec(x) - \tan(x)) = \lim_{x \rightarrow \pi/2} \left(\frac{x + x^3 + \frac{5x^5}{4} + \frac{61x^7}{28} + \ldots}{1 + x^2 + \frac{2x^4}{3} + \frac{34x^6}{105} + \ldots}\right)$$

Simplifying this expression, we get:

\lim_{x \rightarrow \pi/2} (\sec(x) - \tan(x)) = \lim_{x \rightarrow \pi/2} \left(\frac{x + x^