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Magazine Chapter 2: Problem 26
Find the limits. $$\lim _{h \rightarrow 0} \frac{\sin h}{1-\cos h}$$
Short Answer
Expert verified
The limit does not exist; it approaches infinity as \(h\) approaches zero.
Step by step solution
01
Analyze the Limit Expression
First, examine the limit expression: \[\lim _{h \rightarrow 0} \frac{\sin h}{1-\cos h}.\] Notice that directly substituting \(h = 0\) results in the form \(\frac{0}{0}\), which is indeterminate. We need to apply algebraic manipulation to resolve this.
02
Apply Trigonometric Identity
Recall the trigonometric identity:\[ 1 - \cos h = 2 \sin^2 \left( \frac{h}{2} \right).\]Use this substitution in the limit expression:\[\lim _{h \rightarrow 0} \frac{\sin h}{1-\cos h} = \lim _{h \rightarrow 0} \frac{\sin h}{2 \sin^2 \left( \frac{h}{2} \right)}.\]
03
Rewrite Sine with Double Angle Identity
Use the double angle identity for sine:\[\sin h = 2 \sin \left( \frac{h}{2} \right) \cos \left( \frac{h}{2} \right).\]Substitute this into the limit:\[\lim _{h \rightarrow 0} \frac{2 \sin \left( \frac{h}{2} \right) \cos \left( \frac{h}{2} \right)}{2 \sin^2 \left( \frac{h}{2} \right)}.\]
04
Simplify the Expression
Simplify the expression by canceling the common factors:\[\lim _{h \rightarrow 0} \frac{\cos \left( \frac{h}{2} \right)}{\sin \left( \frac{h}{2} \right)}.\]This simplifies to:\[\lim _{h \rightarrow 0} \frac{1}{\tan \left( \frac{h}{2} \right)}.\]
05
Evaluate the Limit
As \(h \to 0\), \( \frac{h}{2} \to 0\). The tangent function behaves such that \(\tan x \approx x\) when \(x\) is near zero. Therefore, evaluate as:\[\lim _{h \rightarrow 0} \frac{1}{\tan \left( \frac{h}{2} \right)} = \lim _{h \rightarrow 0} \frac{1}{\frac{h}{2}} = \frac{2}{h} \to \infty.\]
06
Conclude the Result
The evaluation in Step 5 shows that the limit approaches infinity as \(h\) approaches zero from either side. Thus, the limit does not exist within the framework of finite limits.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Trigonometric Limits
Understanding trigonometric limits is a key skill when it comes to analyzing and evaluating limits that involve trigonometric functions. Trigonometric functions like sine, cosine, and tangent often appear in calculus problems, and their limits can present unique challenges.
When dealing with limits involving trigonometric functions, the values of these functions as the angle approaches zero or other critical points often result in indeterminate forms. To solve these, we use trigonometric identities and other mathematical techniques to simplify and evaluate the limit.
A classic example is evaluating \(\lim_{h \to 0} \frac{\sin h}{1 - \cos h}\). Direct substitution initially results in an indeterminate form, but applying the right identities makes the limit manageable.
When dealing with limits involving trigonometric functions, the values of these functions as the angle approaches zero or other critical points often result in indeterminate forms. To solve these, we use trigonometric identities and other mathematical techniques to simplify and evaluate the limit.
A classic example is evaluating \(\lim_{h \to 0} \frac{\sin h}{1 - \cos h}\). Direct substitution initially results in an indeterminate form, but applying the right identities makes the limit manageable.
Indeterminate Forms
Indeterminate forms occur when directly substituting the limit value into a function results in expressions like \(\frac{0}{0}\). These are situations where the usual arithmetic operations don't apply straightforwardly, and special techniques are required for evaluation.
The exercise presents the limit expression \(\lim_{h \to 0} \frac{\sin h}{1 - \cos h}\), which initially gives the indeterminate form \(\frac{0}{0}\). This tells us that we need to do more work to evaluate this limit properly. By manipulating the expression using algebraic techniques or mathematical identities, we can better understand the behavior of the function as \(h\) approaches zero.
Recognizing indeterminate forms is crucial because they signal us to apply strategies such as L'Hopital's Rule, algebraic manipulation, or trigonometric identities to evaluate the limit.
The exercise presents the limit expression \(\lim_{h \to 0} \frac{\sin h}{1 - \cos h}\), which initially gives the indeterminate form \(\frac{0}{0}\). This tells us that we need to do more work to evaluate this limit properly. By manipulating the expression using algebraic techniques or mathematical identities, we can better understand the behavior of the function as \(h\) approaches zero.
Recognizing indeterminate forms is crucial because they signal us to apply strategies such as L'Hopital's Rule, algebraic manipulation, or trigonometric identities to evaluate the limit.
Trigonometric Identities
Trigonometric identities are mathematical statements that express relationships between the trigonometric functions sine, cosine, tangent, etc. They are crucial tools for simplifying the expressions you'll encounter when solving trigonometric limits.
In the given problem, we start by substituting \(1 - \cos h\) with \(2 \sin^2 \left( \frac{h}{2} \right)\). This identity helps transform the expression into a form that's easier to handle.
Using \(\sin h = 2 \sin \left( \frac{h}{2} \right) \cos \left( \frac{h}{2} \right)\), a double angle identity, also plays a crucial role in manipulating the expression to a non-indeterminate form. These identities are stepping stones that simplify and bring clarity to complex trigonometric expressions, making it easier to evaluate limits.
In the given problem, we start by substituting \(1 - \cos h\) with \(2 \sin^2 \left( \frac{h}{2} \right)\). This identity helps transform the expression into a form that's easier to handle.
Using \(\sin h = 2 \sin \left( \frac{h}{2} \right) \cos \left( \frac{h}{2} \right)\), a double angle identity, also plays a crucial role in manipulating the expression to a non-indeterminate form. These identities are stepping stones that simplify and bring clarity to complex trigonometric expressions, making it easier to evaluate limits.
Double Angle Identities
Double angle identities provide a means to express trigonometric functions of double angles in terms of functions of single angles. They're particularly useful when dealing with expressions that contain multiple occurrences of trigonometric functions like sine and cosine.
In this exercise, we use the double angle identity for sine, which is \(\sin h = 2 \sin \left( \frac{h}{2} \right) \cos \left( \frac{h}{2} \right)\). This substitution simplifies our target expression, allowing us to cancel terms and simplify it into a non-fractional form.
Understanding these identities is essential for simplifying expressions in calculus, especially when the computations involve trigonometric functions, making double angle identities a powerful tool in evaluating limits.
In this exercise, we use the double angle identity for sine, which is \(\sin h = 2 \sin \left( \frac{h}{2} \right) \cos \left( \frac{h}{2} \right)\). This substitution simplifies our target expression, allowing us to cancel terms and simplify it into a non-fractional form.
Understanding these identities is essential for simplifying expressions in calculus, especially when the computations involve trigonometric functions, making double angle identities a powerful tool in evaluating limits.
