Evaluate The Limit: ${ \lim _{x \rightarrow 0} \frac{x^2}{1-\cos X} }$Choose The Correct Answer:(A) -2 (B) 0 (C) 1 (D) 2 (E) Nonexistent

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Introduction

Limits are a fundamental concept in calculus, and evaluating them is crucial in understanding various mathematical concepts. In this article, we will focus on evaluating the limit $\lim_{x \rightarrow 0} \frac{x^2}{1-\cos x}$. This limit is a classic example of a trigonometric limit, and it requires a deep understanding of trigonometric functions and their properties.

Understanding the Limit

The given limit is $\lim_{x \rightarrow 0} \frac{x^2}{1-\cos x}$. To evaluate this limit, we need to understand the behavior of the function as $x$ approaches 0. We can start by analyzing the numerator and denominator separately.

Numerator

The numerator of the function is $x^2$. As $x$ approaches 0, the value of $x^2$ also approaches 0. This is because the square of a small number is even smaller.

Denominator

The denominator of the function is $1-\cos x$. As $x$ approaches 0, the value of $\cos x$ approaches 1. Therefore, the value of $1-\cos x$ approaches 0.

Evaluating the Limit

Now that we have analyzed the numerator and denominator, we can evaluate the limit. We can start by using the fact that $\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1$. This is a well-known trigonometric limit, and it can be proven using various methods.

Using the Trigonometric Limit

We can rewrite the denominator as $1-\cos x = 2\sin^2 \frac{x}{2}$. This is a trigonometric identity, and it can be proven using various methods.

Now, we can substitute this expression into the original limit:

$\lim_{x \rightarrow 0} \frac{x^2}{1-\cos x} = \lim_{x \rightarrow 0} \frac{x^2}{2\sin^2 \frac{x}{2}}$

We can simplify this expression by canceling out the $x^2$ term:

$\lim_{x \rightarrow 0} \frac{x^2}{2\sin^2 \frac{x}{2}} = \lim_{x \rightarrow 0} \frac{1}{2\left(\frac{x}{2}\right)^2 \left(\frac{\sin \frac{x}{2}}{\frac{x}{2}}\right)^2}$

Now, we can use the trigonometric limit $\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1$ to simplify the expression:

$\lim_{x \rightarrow 0} \frac{1}{2\left(\frac{x}{2}\right)^2 \left(\frac{\sin \frac{x}{2}}{\frac{x}{2}}\right)^2} = \lim_{x \rightarrow 0} \frac{1}{2\left(\frac{x}{2}\right)^2 \left(1\right)^2}$

Now, we can simplify the expression by canceling out the $\left(\frac{x}{2}\right)^2$ term:

$\lim_{x \rightarrow 0} \frac{1}{2\left(\frac{x}{2}\right)^2 \left(1\right)^2} = \lim_{x \rightarrow 0} \frac{1}{2\left(\frac{x^2}{4}\right)}$

Now, we can simplify the expression by canceling out the $\frac{1}{2}$ term:

$\lim_{x \rightarrow 0} \frac{1}{2\left(\frac{x^2}{4}\right)} = \lim_{x \rightarrow 0} \frac{4}{x^2}$

Now, we can evaluate the limit by substituting $x = 0$:

$\lim_{x \rightarrow 0} \frac{4}{x^2} = \frac{4}{0^2}$

This expression is undefined, and it approaches infinity.

Conclusion

In this article, we evaluated the limit $\lim_{x \rightarrow 0} \frac{x^2}{1-\cos x}$. We used various trigonometric identities and limits to simplify the expression and evaluate the limit. The final answer is that the limit approaches infinity.

Final Answer

The final answer is: $\boxed{\infty}$

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Introduction

In our previous article, we evaluated the limit $\lim_{x \rightarrow 0} \frac{x^2}{1-\cos x}$. In this article, we will answer some frequently asked questions related to this limit.

Q1: What is the value of the limit $\lim_{x \rightarrow 0} \frac{x^2}{1-\cos x}$?

A1: The value of the limit $\lim_{x \rightarrow 0} \frac{x^2}{1-\cos x}$ is infinity.

Q2: Why is the limit $\lim_{x \rightarrow 0} \frac{x^2}{1-\cos x}$ undefined?

A2: The limit $\lim_{x \rightarrow 0} \frac{x^2}{1-\cos x}$ is undefined because the denominator approaches 0 as $x$ approaches 0.

Q3: Can we simplify the expression $\frac{x^2}{1-\cos x}$?

A3: Yes, we can simplify the expression $\frac{x^2}{1-\cos x}$ by using the trigonometric identity $1-\cos x = 2\sin^2 \frac{x}{2}$.

Q4: How do we evaluate the limit $\lim_{x \rightarrow 0} \frac{x^2}{1-\cos x}$?

A4: We can evaluate the limit $\lim_{x \rightarrow 0} \frac{x^2}{1-\cos x}$ by using the trigonometric limit $\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1$.

Q5: What is the significance of the limit $\lim_{x \rightarrow 0} \frac{x^2}{1-\cos x}$?

A5: The limit $\lim_{x \rightarrow 0} \frac{x^2}{1-\cos x}$ is significant because it is a classic example of a trigonometric limit, and it requires a deep understanding of trigonometric functions and their properties.

Q6: Can we use L'Hopital's rule to evaluate the limit $\lim_{x \rightarrow 0} \frac{x^2}{1-\cos x}$?

A6: No, we cannot use L'Hopital's rule to evaluate the limit $\lim_{x \rightarrow 0} \frac{x^2}{1-\cos x}$ because the limit is undefined.

Q7: What is the relationship between the limit $\lim_{x \rightarrow 0} \frac{x^2}{1-\cos x}$ and the trigonometric function $\cos x$?

A7: The limit $\lim_{x \rightarrow 0} \frac{x^2}{1-\cos x}$ is related to the trigonometric function $\cos x$ because the denominator of the expression is $1-\cos x$.

Q8: Can we use the Taylor series expansion of the trigonometric function $\cos x$ to evaluate the limit $\lim_{x \rightarrow 0} \frac{x^2}{1-\cos x}$?

A8: Yes, we can use the Taylor series expansion of the trigonometric function $\cos x$ to evaluate the limit $\lim_{x \rightarrow 0} \frac{x^2}{1-\cos x}$.

Q9: What is the value of the limit $\lim_{x \rightarrow 0} \frac{x^2}{1-\cos x}$ using the Taylor series expansion of the trigonometric function $\cos x$?

A9: The value of the limit $\lim_{x \rightarrow 0} \frac{x^2}{1-\cos x}$ using the Taylor series expansion of the trigonometric function $\cos x$ is infinity.

Q10: Can we use the limit $\lim_{x \rightarrow 0} \frac{x^2}{1-\cos x}$ to derive other trigonometric limits?

A10: Yes, we can use the limit $\lim_{x \rightarrow 0} \frac{x^2}{1-\cos x}$ to derive other trigonometric limits.

Conclusion

In this article, we answered some frequently asked questions related to the limit $\lim_{x \rightarrow 0} \frac{x^2}{1-\cos x}$. We hope that this article has provided a better understanding of this limit and its significance in trigonometry.