Problem 54
Question
Calculate the given limit. \(\lim _{x \rightarrow 0} \frac{\arcsin (3 x)}{\arctan (2 x)}\)
Step-by-Step Solution
Verified Answer
The limit is \(\frac{3}{2}\).
1Step 1: Evaluate form of the limit
First, substitute \(x = 0\) in both the numerator and the denominator to check if the limit can be directly calculated. We have \(\arcsin(0) = 0\) and \(\arctan(0) = 0\). This gives us the form \(\frac{0}{0}\), which means we need to use L'Hôpital's Rule.
2Step 2: Apply L'Hôpital's Rule
Since we have a \(\frac{0}{0}\) form, we can apply L'Hôpital's Rule. This rule states: if \(\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{0}{0}\), then \(\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}\) (provided the limit on the right exists).
3Step 3: Differentiate the numerator
Differentiate \(\arcsin(3x)\) with respect to \(x\). Using the chain rule: \(\frac{d}{dx}[\arcsin(3x)] = \frac{3}{\sqrt{1-(3x)^2}}\).
4Step 4: Differentiate the denominator
Differentiate \(\arctan(2x)\) with respect to \(x\). Using the chain rule: \(\frac{d}{dx}[\arctan(2x)] = \frac{2}{1+(2x)^2}\).
5Step 5: Evaluate the limit of derivatives
Calculate the limit using the derivatives: \[\lim_{x \to 0} \frac{\frac{3}{\sqrt{1-(3x)^2}}}{\frac{2}{1+(2x)^2}} = \lim_{x \to 0} \frac{3(1+(2x)^2)}{2\sqrt{1-(3x)^2}}.\]Substitute \(x = 0\) again: \[= \frac{3(1+0)}{2\sqrt{1-0}} = \frac{3}{2}.\]
6Step 6: Finalize the result
Hence, the limit \(\lim _{x \rightarrow 0} \frac{\arcsin (3 x)}{\arctan (2 x)}\) is \(\frac{3}{2}\).
Key Concepts
L'Hôpital's RuleLimitDifferentiationChain Rule
L'Hôpital's Rule
L'Hôpital's Rule is a powerful tool in calculus that helps us deal with indeterminate forms like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). When direct substitution of the limit results in these indeterminate forms, L'Hôpital's Rule allows us to differentiate the numerator and denominator separately. If both functions are differentiable, we then take the limit of their derivatives.
To apply L'Hôpital's Rule successfully:
To apply L'Hôpital's Rule successfully:
- Ensure the initial substitution gives an indeterminate form.
- Differentiation of both the numerator and the denominator must be possible.
- Check that the limit of the derivatives provides a determinate result, if not, L'Hôpital's Rule may be applied again or other techniques may be needed.
Limit
In calculus, a limit helps us understand the behavior of a function as it approaches a specific point. It's essentially asking where a function is heading as you get closer and closer to some value along the \(x\)-axis.
The limit symbolically written as \(\lim_{x \to a} f(x) = L\) means that as \(x\) approaches the value \(a\), the function \(f(x)\) approaches \(L\). In our problem, we needed to calculate \(\lim_{x \to 0} \frac{\arcsin(3x)}{\arctan(2x)}\).
Before jumping straight into solving the limit, substituting the value first helps to check if it's a straightforward computation or requires special strategies like L'Hôpital's Rule. This first step is crucial for understanding the type of limit you are dealing with.
The limit symbolically written as \(\lim_{x \to a} f(x) = L\) means that as \(x\) approaches the value \(a\), the function \(f(x)\) approaches \(L\). In our problem, we needed to calculate \(\lim_{x \to 0} \frac{\arcsin(3x)}{\arctan(2x)}\).
Before jumping straight into solving the limit, substituting the value first helps to check if it's a straightforward computation or requires special strategies like L'Hôpital's Rule. This first step is crucial for understanding the type of limit you are dealing with.
Differentiation
Differentiation is one of the core concepts of calculus and is concerned with calculating the derivative of a function. In simpler terms, differentiation is about finding the rate at which a function is changing at any given point.
Derivatives can provide critical insights into the shapes and slopes of curves, which are directly applicable when working with L'Hôpital's Rule.
For our given problem, we differentiated both functions in the fraction:
Derivatives can provide critical insights into the shapes and slopes of curves, which are directly applicable when working with L'Hôpital's Rule.
For our given problem, we differentiated both functions in the fraction:
- \(\arcsin(3x)\)'s derivative with respect to \(x\) is \(\frac{3}{\sqrt{1-(3x)^2}}\).
- \(\arctan(2x)\)'s derivative with respect to \(x\) is \(\frac{2}{1+(2x)^2}\).
Chain Rule
The chain rule is an essential technique in calculus used to differentiate composite functions. It's particularly useful when you have a function within another function, as is often the case with trigonometric expressions.
The chain rule states that if you have a composite function \(f(g(x))\), the derivative is given by \(f'(g(x)) \cdot g'(x)\).
In our example, we applied the chain rule in:
The chain rule states that if you have a composite function \(f(g(x))\), the derivative is given by \(f'(g(x)) \cdot g'(x)\).
In our example, we applied the chain rule in:
- Finding the derivative of \(\arcsin(3x)\), where \(u = 3x\), and applying the formula for the derivative of \(\arcsin(u)\).
- Similarly, for \(\arctan(2x)\), \(u = 2x\), using the derivative formula for \(\arctan(u)\).
Other exercises in this chapter
Problem 54
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