Problem 20
Question
$$ \lim _{x \rightarrow 0} \sin ^{-1}\\{\ln (\cos x)\\} \quad\\{\text { Ans. } 0\\} $$
Step-by-Step Solution
Verified Answer
Considering the limit expression \(\lim_{x \rightarrow 0} \sin^{-1}{\ln (\cos x)}\), we first find the limit of the inner function: \(\lim_{x \rightarrow 0} \ln(\cos x) = \ln(1) = 0\). Since the inner limit is 0, we can apply the limit to the inverse sine function: \(\lim_{x \rightarrow 0} \sin^{-1}{\ln (\cos x)} = \sin^{-1} (0)\), which results in the final limit: \(\lim_{x \rightarrow 0} \sin^{-1}{\ln (\cos x)} = 0\).
1Step 1: Recognize the Inner Function
Since our expression involves composition of functions, first recognize the inner function which in this case is \(\ln(\cos x)\).
2Step 2: Apply L'Hopital's Rule (if needed)
To check if L'Hopital's Rule is necessary, determine the limit of the inner function:
\(\lim_{x \rightarrow 0} \ln(\cos x)\)
As x approaches 0, \(\cos x\) approaches 1. Therefore, the inner limit is:
\(\lim_{x \rightarrow 0} \ln(\cos x) = \ln(1) = 0\)
Since the limit of the inner function is 0, we do not need L'Hopital's Rule.
3Step 3: Apply the limit of inverse sine function
Knowing that the inner limit is 0, we can apply the limit to the inverse sine function:
\(\lim_{x \rightarrow 0} \sin^{-1}{\ln (\cos x)} = \sin^{-1} (0)\)
4Step 4: Calculate the limit
Considering that the inverse sine of 0 is 0, the limit is:
\(\lim_{x \rightarrow 0} \sin^{-1}{\ln (\cos x)} = 0\)
The limit of the given expression as \(x \rightarrow 0\) is 0.
Key Concepts
Understanding l'Hopital's RuleExploring Trigonometric LimitsDealing with Inverse Trigonometric FunctionsUtilizing Natural Logarithm Properties
Understanding l'Hopital's Rule
L'Hopital's Rule is an essential tool in calculus for determining limits that present an indeterminate form, such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). When you encounter a limit of a fraction where both the numerator and the denominator approach zero or infinity, and direct substitution doesn't work, this is when l'Hopital's Rule comes into play. The rule states that if \( \lim_{x \to c} f(x) \) and \( \lim_{x \to c} g(x) \) both approach 0 or both approach \(\pm\infty\), and the derivatives \( f'(x) \) and \( g'(x) \) are continuous near \(c\), then:
\[ \lim_{x \rightarrow c} \frac{f(x)}{g(x)} = \lim_{x \rightarrow c} \frac{f'(x)}{g'(x)} \]
It's a powerful technique for simplifying a difficult limit by differentiating the top and bottom of the fraction separately.
\[ \lim_{x \rightarrow c} \frac{f(x)}{g(x)} = \lim_{x \rightarrow c} \frac{f'(x)}{g'(x)} \]
It's a powerful technique for simplifying a difficult limit by differentiating the top and bottom of the fraction separately.
Exploring Trigonometric Limits
Trigonometric limits are particular limits that involve trigonometric functions such as sine, cosine, or tangent. Understanding their behavior as the variable approaches a certain value is crucial for solving many calculus problems. For example, one fundamental trigonometric limit is:
\[ \lim_{x \rightarrow 0} \frac{\sin x}{x} = 1 \]
This limit is a cornerstone in calculus, and similar properties hold for other trigonometric functions. Keeping these standard limits in mind facilitates the process of solving more complex limits involving trigonometric expressions, as these foundational limits often serve as stepping stones in the solution process.
\[ \lim_{x \rightarrow 0} \frac{\sin x}{x} = 1 \]
This limit is a cornerstone in calculus, and similar properties hold for other trigonometric functions. Keeping these standard limits in mind facilitates the process of solving more complex limits involving trigonometric expressions, as these foundational limits often serve as stepping stones in the solution process.
Dealing with Inverse Trigonometric Functions
Inverse trigonometric functions, such as \(\sin^{-1}(x)\), \(\cos^{-1}(x)\), and \(\tan^{-1}(x)\), are the inverse operations of the basic trigonometric functions. They can often be used to simplify expressions involving arcs, angles, and other trigonometric properties. One should be familiar with their properties and limits, such as:
\[ \lim_{x \rightarrow 0} \sin^{-1}(x) = 0 \]
and
\[ \lim_{x \rightarrow 0} \cos^{-1}(x) = \frac{\pi}{2} \]
These properties can be extremely useful in calculating limits that involve composite functions with an inverse trigonometric function.
\[ \lim_{x \rightarrow 0} \sin^{-1}(x) = 0 \]
and
\[ \lim_{x \rightarrow 0} \cos^{-1}(x) = \frac{\pi}{2} \]
These properties can be extremely useful in calculating limits that involve composite functions with an inverse trigonometric function.
Utilizing Natural Logarithm Properties
The natural logarithm, represented by \(\ln(x)\), has unique properties that are pivotal in solving calculus problems, especially when dealing with limits. One of the most fundamental properties is that the natural logarithm of 1 is 0:
\[ \ln(1) = 0 \]
Additionally, the function \(\ln(x)\) is continuous and differentiable for all \(x > 0\), which allows us to apply various calculus techniques, like differentiation and integration, with ease. When analyzing limits involving natural logarithms, these properties simplify the process and lead to a clearer understanding of the limit's behavior.
\[ \ln(1) = 0 \]
Additionally, the function \(\ln(x)\) is continuous and differentiable for all \(x > 0\), which allows us to apply various calculus techniques, like differentiation and integration, with ease. When analyzing limits involving natural logarithms, these properties simplify the process and lead to a clearer understanding of the limit's behavior.
Other exercises in this chapter
Problem 18
$$ \lim _{x \rightarrow 1} \cos ^{-1} x \cdot \ln x+\frac{1}{x}+\frac{1}{x^{2}}\\{\text { Ans. } 2\\} $$
View solution Problem 19
$$ \lim _{x \rightarrow 1} \ln (\sin \pi x-\cos \pi x) \quad\\{\text { Ans. } 0\\} $$
View solution Problem 21
$$ \lim _{x \rightarrow 0} \operatorname{sgn}\\{\ln (\cos x)\\}\\{\text { Ans. }-1\\} $$
View solution Problem 22
$$ \lim _{x \rightarrow 0} \operatorname{sgn}\\{\ln (1+x)\\}\\{\text { Ans. } 1,-1\\} $$
View solution