Problem 53
Question
Find the limits in Problems Be sure to check whether you can apply I'Hospital's rule before you evaluate the limit. $$ \lim _{x \rightarrow(\pi / 2)^{-}}(\tan x+\sec x) $$
Step-by-Step Solution
Verified Answer
The limit is \(+\infty\).
1Step 1: Understanding the Limit
We have the limit \( \lim _{x \rightarrow(\pi / 2)^{-}}(\tan x+\sec x) \). We need to find the behavior of the function as \( x \) approaches \( \pi/2 \) from the left.
2Step 2: Properties of Trigonometric Functions
As \( x \) approaches \( \pi/2 \) from the left, \( \tan x \rightarrow +\infty \) and \( \sec x = \frac{1}{\cos x} \rightarrow +\infty \) since \( \cos x \rightarrow 0^{+} \).
3Step 3: Checking Applicability of L'Hospital's Rule
L'Hospital's Rule is used for indeterminate forms such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). Since both \( \tan x \) and \( \sec x \) tend towards positive infinity, the expression tends towards \(+\infty + +\infty = +\infty\). Thus, L'Hospital's Rule is not applicable.
4Step 4: Conclusion on the Limit
Since both terms \( \tan x \) and \( \sec x \) approach \(+\infty\) as \( x \to (\pi/2)^{-} \), the sum also approaches \(+\infty\). Therefore, the limit is \(+\infty\).
Key Concepts
Trigonometric FunctionsL'Hospital's RuleIndeterminate Forms
Trigonometric Functions
Trigonometric functions are core concepts in calculus and are essential for evaluating limits, especially when dealing with angles and periodic behaviors. Two such functions are the tangent and secant functions. The tangent, denoted as \( \tan x \), is the ratio of the sine to the cosine of \( x \). The secant, written as \( \sec x \), is the reciprocal of the cosine i.e., \( \sec x = \frac{1}{\cos x} \). Both of these functions have interesting behaviors as certain angles are approached.
- The tangent function tends towards \( +\infty \) as \( x \) approaches \( \frac{\pi}{2} \) from the left, since \( \tan x = \frac{\sin x}{\cos x} \) and \( \cos x \) approaches zero from the positive side.
- The secant function also approaches \( +\infty \) in this scenario, because the cosine approaches zero, causing its reciprocal \( 1/\cos x \) to shoot up towards infinity.
L'Hospital's Rule
L'Hospital's Rule is a powerful technique in calculus used to evaluate limits resulting in indeterminate forms. Typically, these forms are \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). When you encounter such forms, you can use L'Hospital's Rule by differentiating the numerator and the denominator separately and then take the limit of that new fraction. However, not all limits qualify for this treatment.
For the limit \( \lim _{x \rightarrow(\pi / 2)^{-}}(\tan x+\sec x) \), L'Hospital's Rule is not applicable. This is because the expression \( \tan x + \sec x \) approaches \( +\infty \) directly, not through a fraction that could simplify using L'Hospital's Rule. It is only applicable when both the numerator and denominator of a fraction individually approach zero or infinity, and that's not the case here.
This illustrates an important point: always check the form of your limit before defaulting to L'Hospital's Rule, as it might not be possible or necessary to apply it, depending on the situation.
For the limit \( \lim _{x \rightarrow(\pi / 2)^{-}}(\tan x+\sec x) \), L'Hospital's Rule is not applicable. This is because the expression \( \tan x + \sec x \) approaches \( +\infty \) directly, not through a fraction that could simplify using L'Hospital's Rule. It is only applicable when both the numerator and denominator of a fraction individually approach zero or infinity, and that's not the case here.
This illustrates an important point: always check the form of your limit before defaulting to L'Hospital's Rule, as it might not be possible or necessary to apply it, depending on the situation.
Indeterminate Forms
Indeterminate forms often arise in calculus when evaluating certain limits, and can be tricky to handle. They include forms like \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), and other expressions such as \( \infty - \infty \). When encountering these, they indicate that the situation is ambiguous or not straightforward, requiring further examination or techniques to resolve.
In the given exercise, the expression \( \tan x + \sec x \) does not form an indeterminate like \( \frac{\infty}{\infty} \) since it's directly a sum of two quantities tending independently to \( +\infty \). Such scenarios can often be resolved by recognizing their behavior instead, like knowing that the growth of each term towards infinity solidifies the result of the limit.
Understanding the nature of the limit and how different components interact is crucial for effectively evaluating limits, especially when indeterminate forms may seem apparent but don't actually exist.
In the given exercise, the expression \( \tan x + \sec x \) does not form an indeterminate like \( \frac{\infty}{\infty} \) since it's directly a sum of two quantities tending independently to \( +\infty \). Such scenarios can often be resolved by recognizing their behavior instead, like knowing that the growth of each term towards infinity solidifies the result of the limit.
Understanding the nature of the limit and how different components interact is crucial for effectively evaluating limits, especially when indeterminate forms may seem apparent but don't actually exist.
Other exercises in this chapter
Problem 52
In Problems 47-58, find the general solution of the differential equation. $$ \frac{d y}{d t}=t^{2}\left(1-t^{2}\right), t \geq 0 $$
View solution Problem 52
Suppose that \(f\) is differentiable for all \(x \in \mathbf{R}\) and, furthermore, that \(f\) satisfies \(f(0)=0\) and \(1 \leq f^{\prime}(x) \leq 2\) for all
View solution Problem 53
In Problems 47-58, find the general solution of the differential equation. $$ \frac{d y}{d t}=e^{-t / 2}, t \geq 0 $$
View solution Problem 53
Suppose that \(f\) is differentiable for all \(x \in \mathbf{R}\) with \(f(2)=3\) and \(f^{\prime}(x)=0\) for all \(x \in \mathbf{R}\). Find \(f(x)\).
View solution