Chapter 2: Problem 301
$$ \lim _{x \rightarrow 0}(\cos x+a \sin b x)^{\frac{1}{x}}\left\\{\text { Ans. } e^{a b}\right\\} $$
Short Answer
Expert verified
The short answer for the given limit problem is the value of the limit, which is: \(\lim _{x \rightarrow 0}(\cos x+a \sin b x)^{\frac{1}{x}} = e^{ab}\).
Step by step solution
01
1. Rewrite the expression in exponential form
The given expression is an exponential function. We can rewrite the limit expression in exponential form for simplification: \(\lim _{x \rightarrow 0}e^{\frac{1}{x}\ln(\cos x+a \sin b x)}\)
02
2. Apply L'Hopital's Rule
We now need to find the derivative of the exponent, \(\frac{1}{x}\ln(\cos x+a \sin b x)\), with respect to x. L'Hopital's rule requires having the form \(\frac{f(x)}{g(x)}\). To apply L'Hopital's Rule, let's rewrite the expression as follows: \(\lim_{x \rightarrow 0} e^{\frac{\ln(\cos x+a \sin b x)}{x}}\)
Now, applying L'Hopital's Rule, we need to find the derivative of numerator and the derivative of denominator with respect to x:
Numerator derivative: \(\frac{d}{dx} [\ln(\cos x+a \sin b x)] = \frac{-\sin x + ab \cos bx}{\cos x+a \sin b x}\)
Denominator derivative: \(\frac{d}{dx} [x] = 1\)
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3. Compute the limit
Now that we have the derivatives of the numerator and denominator, let's compute the limit:
\(\lim_{x \rightarrow 0} e^{\frac{\frac{-\sin x + ab \cos bx}{\cos x+a \sin b x}}{1}} = e^{\frac{-\sin x + ab \cos bx}{\cos x+a \sin b x}}\)
As \(x \rightarrow 0\), the expression simplifies to:
\(e^{\frac{-(0) + ab \cos b(0)}{\cos (0)+a \sin b(0)}} = e^{\frac{ab \cdot 1}{1 + a \cdot 0}} = e^{ab}\)
Therefore, the value of the limit is:
\(\lim _{x \rightarrow 0}(\cos x+a \sin b x)^{\frac{1}{x}} = e^{ab}\)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
L'Hopital's Rule
L'Hopital's Rule is a powerful tool in calculus used to determine limits that have indeterminate forms, such as \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). When you encounter a limit with these indeterminate forms, you can apply L'Hopital's Rule to simplify the calculation. Let's break down how it works:
- If \(\lim_{x \to c} f(x) = 0\) and \(\lim_{x \to c} g(x) = 0\), or both approach infinity, then:
- Here, \(f'(x)\) and \(g'(x)\) are the derivatives of \(f(x)\) and \(g(x)\) respectively.
- If the original limit still results in an indeterminate form, apply L'Hopital's Rule again.
- L'Hopital's Rule works when both functions involved are differentiable around the point of interest.
Exponential Functions
Exponential functions are fundamental in calculus and mathematics. They have a form like \(e^x\), where \(e\) is the base of the natural logarithm, approximately equal to 2.718. These functions grow rapidly and appear in real-world applications like population growth, radioactive decay, and interest calculations.
When dealing with limits involving exponential functions, especially those in the form \(e^{f(x)}\), it’s helpful to use logarithms to simplify the expression. In our problem, the expression \((\cos x + a \sin bx)^{\frac{1}{x}}\) was rewritten as an exponential function, making use of the identity:
When dealing with limits involving exponential functions, especially those in the form \(e^{f(x)}\), it’s helpful to use logarithms to simplify the expression. In our problem, the expression \((\cos x + a \sin bx)^{\frac{1}{x}}\) was rewritten as an exponential function, making use of the identity:
- \((...)= e^{\ln(...)},\)
- which allows us to transform the expression into \(e^{\frac{1}{x} \ln(\cos x + a \sin bx)}\).
Trigonometric Functions
Trigonometric functions such as sine and cosine are paramount in calculus. They describe periodic phenomena and are crucial in a variety of applications ranging from physics to engineering.
In our problem, we use the trigonometric functions \(\cos x\) and \(\sin bx\) inside a limit. These functions help illustrate:
In our problem, we use the trigonometric functions \(\cos x\) and \(\sin bx\) inside a limit. These functions help illustrate:
- The interplay between oscillation (sine function) and rotation (cosine function).
- The importance of having a handle on their derivatives: \(\cos x\) differentiates to \(-\sin x\), and \(\sin bx\) differentiates to \(b \cos bx\).