Use The Trigonometric Limit $\lim_{x \rightarrow 0} \frac{\sin X}{x} = 1$ And/or $\lim_{x \rightarrow 0} \frac{\cos X - 1}{x} = 0$ To Evaluate The Following Limit:$\lim_{x \rightarrow 0} \frac{\sin 5x}{19x}$Select The Correct

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Introduction

In mathematics, trigonometric limits are a fundamental concept that deals with the behavior of trigonometric functions as the input values approach a specific point, often zero. These limits are crucial in calculus and are used to evaluate various types of limits, including indeterminate forms. In this article, we will explore how to use the trigonometric limits limx0sinxx=1\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1 and limx0cosx1x=0\lim_{x \rightarrow 0} \frac{\cos x - 1}{x} = 0 to evaluate the limit limx0sin5x19x\lim_{x \rightarrow 0} \frac{\sin 5x}{19x}.

Understanding the Trigonometric Limits

Before we dive into the evaluation of the given limit, let's first understand the two trigonometric limits that we will be using.

Limit 1: limx0sinxx=1\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1

The limit limx0sinxx=1\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1 is a fundamental limit in calculus that deals with the behavior of the sine function as the input value approaches zero. This limit is often used to evaluate various types of limits, including indeterminate forms.

Limit 2: limx0cosx1x=0\lim_{x \rightarrow 0} \frac{\cos x - 1}{x} = 0

The limit limx0cosx1x=0\lim_{x \rightarrow 0} \frac{\cos x - 1}{x} = 0 is another fundamental limit in calculus that deals with the behavior of the cosine function as the input value approaches zero. This limit is often used to evaluate various types of limits, including indeterminate forms.

Evaluating the Given Limit

Now that we have understood the two trigonometric limits, let's use them to evaluate the given limit limx0sin5x19x\lim_{x \rightarrow 0} \frac{\sin 5x}{19x}.

Step 1: Rewrite the Limit

To evaluate the given limit, we can start by rewriting it in a form that involves the two trigonometric limits that we have learned.

limx0sin5x19x=limx0sin5x5x119\lim_{x \rightarrow 0} \frac{\sin 5x}{19x} = \lim_{x \rightarrow 0} \frac{\sin 5x}{5x} \cdot \frac{1}{19}

Step 2: Apply the Trigonometric Limits

Now that we have rewritten the limit, we can apply the two trigonometric limits that we have learned.

limx0sin5x5x119=limx0sin5x5xlimx0119\lim_{x \rightarrow 0} \frac{\sin 5x}{5x} \cdot \frac{1}{19} = \lim_{x \rightarrow 0} \frac{\sin 5x}{5x} \cdot \lim_{x \rightarrow 0} \frac{1}{19}

Step 3: Evaluate the Limits

Now that we have applied the two trigonometric limits, we can evaluate the limits.

limx0sin5x5x=1\lim_{x \rightarrow 0} \frac{\sin 5x}{5x} = 1 (by Limit 1)

limx0119=119\lim_{x \rightarrow 0} \frac{1}{19} = \frac{1}{19}

Step 4: Combine the Results

Now that we have evaluated the limits, we can combine the results to get the final answer.

limx0sin5x19x=1119=119\lim_{x \rightarrow 0} \frac{\sin 5x}{19x} = 1 \cdot \frac{1}{19} = \frac{1}{19}

Conclusion

In this article, we have used the trigonometric limits limx0sinxx=1\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1 and limx0cosx1x=0\lim_{x \rightarrow 0} \frac{\cos x - 1}{x} = 0 to evaluate the limit limx0sin5x19x\lim_{x \rightarrow 0} \frac{\sin 5x}{19x}. We have shown that the limit evaluates to 119\frac{1}{19}.

Final Answer

The final answer is 119\boxed{\frac{1}{19}}.

Additional Examples

Here are some additional examples of how to use the trigonometric limits to evaluate various types of limits.

Example 1: limx0sin3x6x\lim_{x \rightarrow 0} \frac{\sin 3x}{6x}

To evaluate this limit, we can use the trigonometric limit limx0sinxx=1\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1.

limx0sin3x6x=limx0sin3x3x12=112=12\lim_{x \rightarrow 0} \frac{\sin 3x}{6x} = \lim_{x \rightarrow 0} \frac{\sin 3x}{3x} \cdot \frac{1}{2} = 1 \cdot \frac{1}{2} = \frac{1}{2}

Example 2: limx0cos2x12x\lim_{x \rightarrow 0} \frac{\cos 2x - 1}{2x}

To evaluate this limit, we can use the trigonometric limit limx0cosx1x=0\lim_{x \rightarrow 0} \frac{\cos x - 1}{x} = 0.

limx0cos2x12x=limx0cos2x12x=0\lim_{x \rightarrow 0} \frac{\cos 2x - 1}{2x} = \lim_{x \rightarrow 0} \frac{\cos 2x - 1}{2x} = 0

Conclusion

In this article, we have shown how to use the trigonometric limits limx0sinxx=1\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1 and limx0cosx1x=0\lim_{x \rightarrow 0} \frac{\cos x - 1}{x} = 0 to evaluate various types of limits. We have provided examples of how to use these limits to evaluate limits involving sine and cosine functions.

Introduction

In our previous article, we explored how to use the trigonometric limits limx0sinxx=1\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1 and limx0cosx1x=0\lim_{x \rightarrow 0} \frac{\cos x - 1}{x} = 0 to evaluate various types of limits. In this article, we will answer some of the most frequently asked questions about trigonometric limits.

Q: What is the significance of the trigonometric limits?

A: The trigonometric limits are fundamental in calculus and are used to evaluate various types of limits, including indeterminate forms. They are also used to derive various mathematical formulas and theorems.

Q: How do I remember the trigonometric limits?

A: One way to remember the trigonometric limits is to use the following mnemonic device: "Sine over X equals 1, Cosine minus 1 over X equals 0". This can help you remember the two limits.

Q: Can I use the trigonometric limits to evaluate limits involving other trigonometric functions?

A: Yes, you can use the trigonometric limits to evaluate limits involving other trigonometric functions. For example, you can use the limit limx0sinxx=1\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1 to evaluate the limit limx0tanxx\lim_{x \rightarrow 0} \frac{\tan x}{x}.

Q: How do I apply the trigonometric limits to evaluate limits involving multiple trigonometric functions?

A: To apply the trigonometric limits to evaluate limits involving multiple trigonometric functions, you can use the following steps:

  1. Rewrite the limit in a form that involves the trigonometric functions.
  2. Apply the trigonometric limits to the rewritten limit.
  3. Simplify the resulting expression to get the final answer.

Q: Can I use the trigonometric limits to evaluate limits involving non-trigonometric functions?

A: No, the trigonometric limits are only applicable to limits involving trigonometric functions. However, you can use other mathematical techniques, such as L'Hopital's rule, to evaluate limits involving non-trigonometric functions.

Q: Are there any other trigonometric limits that I should know about?

A: Yes, there are several other trigonometric limits that you should know about, including:

  • limx0tanxx=1\lim_{x \rightarrow 0} \frac{\tan x}{x} = 1
  • limx0cotxx=1\lim_{x \rightarrow 0} \frac{\cot x}{x} = 1
  • limx0secx1x=0\lim_{x \rightarrow 0} \frac{\sec x - 1}{x} = 0
  • limx0cscx1x=0\lim_{x \rightarrow 0} \frac{\csc x - 1}{x} = 0

Q: How do I practice using the trigonometric limits?

A: To practice using the trigonometric limits, you can try evaluating various limits involving trigonometric functions. You can also use online resources, such as limit calculators, to help you evaluate limits.

Conclusion

In this article, we have answered some of the most frequently asked questions about trigonometric limits. We have provided examples of how to use the trigonometric limits to evaluate various types of limits and have discussed some of the most common applications of these limits.

Additional Resources

Here are some additional resources that you can use to learn more about trigonometric limits:

  • Limit Calculators: Online limit calculators can help you evaluate limits involving trigonometric functions.
  • Mathematical Textbooks: Mathematical textbooks, such as "Calculus" by Michael Spivak, provide a comprehensive introduction to trigonometric limits.
  • Online Courses: Online courses, such as "Calculus" on Coursera, provide a comprehensive introduction to trigonometric limits.

Final Answer

The final answer is 1\boxed{1}.

Final Example

Here is a final example of how to use the trigonometric limits to evaluate a limit:

limx0sin2x3x\lim_{x \rightarrow 0} \frac{\sin 2x}{3x}

To evaluate this limit, we can use the trigonometric limit limx0sinxx=1\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1.

limx0sin2x3x=limx0sin2x2x13=113=13\lim_{x \rightarrow 0} \frac{\sin 2x}{3x} = \lim_{x \rightarrow 0} \frac{\sin 2x}{2x} \cdot \frac{1}{3} = 1 \cdot \frac{1}{3} = \frac{1}{3}