Problem 116
Question
$$ \lim _{x \rightarrow 0} \frac{\sin \alpha x}{\sin \beta x}\left\\{\text { Ans. } \frac{a}{\beta}\right\\} $$
Step-by-Step Solution
Verified Answer
The solution to the limit problem is \(\frac{\alpha}{\beta}\).
1Step 1: Applying L'Hopital’s Rule
Since the limit takes the form \(\frac{0}{0}\) at \(x = 0\), l'Hopital's rule can be used, which states that \(\lim_{x\rightarrow c}\frac{f(x)}{g(x)} = \lim_{x\rightarrow c}\frac{f'(x)}{g'(x)}\), where \(f(x)\) and \(g(x)\) represent differentiable functions. Deriving the numerator and denominator separately, we have \(\frac{d}{dx}\sin(\alpha x) = \alpha cos(\alpha x)\) and \(\frac{d}{dx}\sin(\beta x) = \beta cos(\beta x)\).
2Step 2: Substituting The Derivatives
Now, replace the original quotient with the derivative of the numerator divided by derivative of denominator, per l'Hopital's rule. This gives \(\lim_{x\rightarrow 0}\frac{\alpha cos(\alpha x)}{\beta cos(\beta x)}\).
3Step 3: Simplifying the Expression
At \(x = 0\), \(cos(\alpha x) = cos(0) = 1\) and \(cos(\beta x) = cos(0) = 1\). So, the limit simplifies to \(\lim_{x\rightarrow 0}\frac{\alpha}{\beta} = \frac{\alpha}{\beta}\).
Key Concepts
Limits in CalculusTrigonometric LimitsDifferentiable Functions
Limits in Calculus
Understanding limits is a foundational concept in calculus. Limits help us determine the value that a function approaches as the input gets infinitely close to a point. It's a way of analyzing the behavior of functions at the edge of their domains.
In the case of the given problem, we approach the limit as the variable, denoted by x, approaches 0. The expression \( \frac{\sin \alpha x}{\sin \beta x} \) initially presents a challenge because directly substituting 0 for x gives us an indeterminate form of \(\frac{0}{0}\). This is precisely the situation where the concept of limits shines, as we need to figure out what value the expression is approaching, rather than what it equals at the point.
In the case of the given problem, we approach the limit as the variable, denoted by x, approaches 0. The expression \( \frac{\sin \alpha x}{\sin \beta x} \) initially presents a challenge because directly substituting 0 for x gives us an indeterminate form of \(\frac{0}{0}\). This is precisely the situation where the concept of limits shines, as we need to figure out what value the expression is approaching, rather than what it equals at the point.
Trigonometric Limits
Trigonometric limits involve functions like sine and cosine, which frequently appear in calculus problems. The behavior of these trigonometric functions around certain points, often at 0 or \(\pi\), can lead to the indeterminate forms we discussed in the limits section.
For solving trigonometric limits, we can often use standard limits such as \(\lim_{x\rightarrow 0}\frac{\sin(x)}{x} = 1\), or like in our example, apply L'Hopital's Rule when faced with an indeterminate form. This rule is often a last resort specifically reserved for situations where more straightforward methods do not readily present a solution, and our function is differentiable at the point in question.
For solving trigonometric limits, we can often use standard limits such as \(\lim_{x\rightarrow 0}\frac{\sin(x)}{x} = 1\), or like in our example, apply L'Hopital's Rule when faced with an indeterminate form. This rule is often a last resort specifically reserved for situations where more straightforward methods do not readily present a solution, and our function is differentiable at the point in question.
Differentiable Functions
Differentiable functions are those that have a derivative at each point in their domain. In simpler terms, if you can draw a tangent line at every point along the curve of a function, it's differentiable. This concept is crucial for applying L'Hopital's Rule, as we do only with differentiable functions.
In the textbook exercise, we have \(\sin(\alpha x)\) and \(\sin(\beta x)\), both of which are differentiable. The derivatives, \(\alpha \cos(\alpha x)\) and \(\beta \cos(\beta x)\), replace the original functions in the numerator and denominator, respectively. This step is critical to the process—without the ability to differentiate these functions, we wouldn't be able to apply L'Hopital’s Rule and thus solve the limit.
In the textbook exercise, we have \(\sin(\alpha x)\) and \(\sin(\beta x)\), both of which are differentiable. The derivatives, \(\alpha \cos(\alpha x)\) and \(\beta \cos(\beta x)\), replace the original functions in the numerator and denominator, respectively. This step is critical to the process—without the ability to differentiate these functions, we wouldn't be able to apply L'Hopital’s Rule and thus solve the limit.
Other exercises in this chapter
Problem 114
$$ \lim _{x \rightarrow 0} \frac{\sin 3 x}{x}\\{\text { Ans. } 3\\} $$
View solution Problem 115
$$ \text { 115. } \lim _{x \rightarrow 0} \frac{\tan k x}{x}\\{\text { Ans. } k\\} $$
View solution Problem 117
$$ \left.\lim _{x \rightarrow 0} \frac{\tan 2 x}{\sin 5 x} \text { \\{Ans. } \frac{2}{5}\right\\} $$
View solution Problem 118
$$ \lim _{x \rightarrow 1} \frac{\sin (\ln x)}{\ln x}\\{\text { Ans. } 1\\} $$
View solution