Problem 209
Question
$$ \lim _{x \rightarrow 1}(1+\sin \pi x)^{\cot \pi x}\left\\{\text { Ans. } \frac{1}{c}\right\\} $$
Step-by-Step Solution
Verified Answer
The short answer is: \(\lim_{x \rightarrow 1}(1+\sin \pi x)^{\cot \pi x} = e\).
1Step 1: Rewrite the Limit Expression using Natural Logarithm
To make it easier to apply L'Hôpital's rule, first rewrite the given expression using natural logarithm. Associate the limit with the power as \(y\), then take the natural logarithm of both sides:
\[y = (1+\sin \pi x)^{\cot \pi x}\]
\[\ln(y) = \ln((1+\sin \pi x)^{\cot \pi x}) = \cot \pi x \cdot \ln(1+\sin \pi x)\]
Now find the limit of the new expression:
\[\lim_{x \rightarrow 1} \ln(y) = \lim_{x \rightarrow 1} (\cot \pi x \cdot \ln(1+\sin \pi x))\]
2Step 2: Rewrite the Limit as a Fraction
Rewrite the limit as a fraction in order to apply L'Hôpital's rule:
\[\lim_{x \rightarrow 1} \frac{\ln(1+\sin \pi x)}{\frac{1}{\cot\pi x}}\]
3Step 3: Apply L'Hôpital's Rule
Calculate the derivatives of the numerator and the denominator:
**Numerator:**
\[\frac{d}{dx}(\ln(1+\sin \pi x)) = \frac{(\pi \cos \pi x)}{1+\sin \pi x}\]
**Denominator:**
\[\frac{d}{dx}\frac{1}{\cot \pi x} = \frac{-\pi \csc^2 \pi x}{1}\]
Now write the limit with the derivatives of the numerator and the denominator and apply L'Hôpital's rule:
\[\lim_{x \rightarrow 1} \frac{\frac{(\pi \cos \pi x)}{1+\sin \pi x}}{-\pi \csc^2 \pi x}\]
It's now possible to evaluate this limit directly:
\[\frac{\frac{(\pi \cos \pi )}{1+\sin \pi }}{-\pi \csc^2 \pi }\]
Since \(\cos(\pi) = -1\), \(\sin(\pi) = 0\), and \(\csc(\pi) = \frac{1}{\sin(\pi)}\), the limit simplifies to:
\[\frac{\frac{\pi}{1}}{\pi}\]
Which further simplifies to:
\[\lim_{x \rightarrow 1} \ln(y) = 1\]
4Step 4: Solve for the Limit of the Original Expression
To find the original limit, remove the natural logarithm from the limit:
\[y = e^1 = e\]
Thereby, the answer of the given exercise is:
\[\lim_{x \rightarrow 1}(1+\sin \pi x)^{\cot \pi x} = e\]
Key Concepts
Limit EvaluationNatural LogarithmDerivativesTrigonometric Functions
Limit Evaluation
Understanding how to evaluate limits is crucial when working with continuous functions. Limits represent the value that a function approaches as the input (or index) approaches some value. In this particular problem, the goal is to find the limit of the expression \(\lim _{x \rightarrow 1}(1+\sin \pi x)^{\cot \pi x}\) as \(x\) approaches 1.
Evaluating this directly was complex, so we used a simplification technique involving the natural logarithm. Taking the logarithm allows us to convert the original expression into a form where L'Hôpital's Rule becomes applicable.
By converting the expression into a logarithmic form, we transformed the problem into evaluating a limit of a simpler expression \(\lim_{x \rightarrow 1} \ln((1+\sin \pi x)^{\cot \pi x})\).
This transformation is valuable because it enables the application of L'Hôpital's Rule, which is used to solve indeterminate forms like \(0/0\) or \(\infty/\infty\). This technique simplifies solving such complex limits.
Evaluating this directly was complex, so we used a simplification technique involving the natural logarithm. Taking the logarithm allows us to convert the original expression into a form where L'Hôpital's Rule becomes applicable.
By converting the expression into a logarithmic form, we transformed the problem into evaluating a limit of a simpler expression \(\lim_{x \rightarrow 1} \ln((1+\sin \pi x)^{\cot \pi x})\).
This transformation is valuable because it enables the application of L'Hôpital's Rule, which is used to solve indeterminate forms like \(0/0\) or \(\infty/\infty\). This technique simplifies solving such complex limits.
Natural Logarithm
The natural logarithm is frequently used in calculus to simplify expressions, allowing us to manipulate exponential expressions by converting exponents into multiplications. It is denoted by \(\ln\) and is particularly useful because it is the inverse of the exponential function.
In the original exercise, we used the natural logarithm to simplify the expression \((1+\sin \pi x)^{\cot \pi x}\) by transforming it into:\[ \ln((1+\sin \pi x)^{\cot \pi x}) = \cot \pi x \cdot \ln(1+\sin \pi x). \]
This transformation makes complex calculus problems easier to handle, especially using tools like L'Hôpital's Rule. The logarithmic transformation is essential in reducing the computation complexity by changing multiplicative relationships into additive ones, which are easier to differentiate.
Understanding how to intelligently use natural logarithms in calculus enables learners to tackle a wide variety of limit evaluation problems.
In the original exercise, we used the natural logarithm to simplify the expression \((1+\sin \pi x)^{\cot \pi x}\) by transforming it into:\[ \ln((1+\sin \pi x)^{\cot \pi x}) = \cot \pi x \cdot \ln(1+\sin \pi x). \]
This transformation makes complex calculus problems easier to handle, especially using tools like L'Hôpital's Rule. The logarithmic transformation is essential in reducing the computation complexity by changing multiplicative relationships into additive ones, which are easier to differentiate.
Understanding how to intelligently use natural logarithms in calculus enables learners to tackle a wide variety of limit evaluation problems.
Derivatives
Derivatives measure how a function changes as its input changes and are central to calculus. They provide the necessary information to perform operations like L'Hôpital's Rule. In this exercise, derivative calculation was critical for evaluating the limit.
To apply L'Hôpital's Rule, computing the derivatives of the numerator and the denominator of our transformed expression was necessary. The expressions were:
To apply L'Hôpital's Rule, computing the derivatives of the numerator and the denominator of our transformed expression was necessary. The expressions were:
- Numerator: \( \ln(1+\sin \pi x) \rightarrow \frac{(\pi \cos \pi x)}{1+\sin \pi x} \)
- Denominator: \( \frac{1}{\cot \pi x} \rightarrow -\pi \csc^2 \pi x \)
Trigonometric Functions
Trigonometric functions like \(\sin\), \(\cos\), \(\tan\), \(\cot\), and \(\csc\) are vital in various fields of mathematics, including calculus. In this exercise, they play a central role.
The exercise involved the function \((1+\sin \pi x)^{\cot \pi x}\), which incorporates the sine and cotangent trigonometric functions. These functions are periodic and have specific values at key points like \(\pi\), which influence the limit's evaluation. For instance, at \(x = 1\):
By knowing the properties and special values of trigonometric functions, students can more effectively tackle calculus problems like the one presented in this exercise.
The exercise involved the function \((1+\sin \pi x)^{\cot \pi x}\), which incorporates the sine and cotangent trigonometric functions. These functions are periodic and have specific values at key points like \(\pi\), which influence the limit's evaluation. For instance, at \(x = 1\):
- \(\sin(\pi x) = \sin(\pi) = 0\)
- \(\cos(\pi x) = \cos(\pi) = -1\)
- \(\csc(\pi x) = \frac{1}{\sin(\pi x)}\)
By knowing the properties and special values of trigonometric functions, students can more effectively tackle calculus problems like the one presented in this exercise.
Other exercises in this chapter
Problem 207
$$ \lim _{x \rightarrow 3} \frac{x-3}{|x-3|} \text { . }\\{\text { Ans. } 1,-1\\} $$
View solution Problem 208
$$ \left.\lim _{x \rightarrow 0}\left(\frac{1+\tan x}{1+\sin x}\right)^{\frac{1}{3 x}} \text { \\{Ans. } 1\right\\} $$
View solution Problem 210
$$ \lim _{x \rightarrow 0} \frac{\cos x+4 \tan x}{2-x-2 x^{4}}\left\\{\text { Ans, } \frac{1}{2}\right\\} $$
View solution Problem 211
$$ \lim _{x \rightarrow \infty} \frac{1-2 x}{\sqrt[3]{1+8 x^{3}}}+2^{-x^{2}}\\{\text { Ans. }-1\\} $$
View solution