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Let$$g(x)=\left\\{\begin{array}{ll}\frac{1-\cos x}{2 x} & \text { if } x \neq 0 \\\a & \text { if } x=0\end{array}\right.$$ For what values of \(a\) is \(g\) continuous?

Short Answer

Expert verified
Based on the step-by-step solution provided: Question: Determine the value of \(a\) for which the function \(g(x)\) is continuous, if \(g(x)\) is defined as: \(g(x) =\begin{cases} \frac{1-\cos x}{2x}, & \text{if } x\neq 0,\\ a, & \text{if } x=0. \end{cases}\) Answer: To make \(g(x)\) continuous, it must be the case that \(g(0) = a = 0\).

Step by step solution

01

Find the limit of the numerator as x approaches 0

To find the limit of the given function as \(x\) approaches \(0\), we first need to look at the behavior of the numerator \(1-\cos x\) as \(x\) approaches \(0\). We already know that the limit of \(\cos x\) as \(x\) approaches \(0\) is \(1\).
02

Using the solution of step 1 to analyze the function

Using the fact that the numerator approaches \(0\) and the limit of the denominator (\(2x\)) approaches \(0\) as \(x\) approaches \(0\), we have a \(\frac{0}{0}\) form which is an indeterminate form. In order to find the limit, we can use L'Hôpital's rule.
03

Applying L'Hôpital's rule

According to L'Hôpital's rule, \(\lim _{x \to a} \frac{f(x)}{g(x)}=\lim _{x \to a} \frac{f'(x)}{g'(x)}\) if the limit of the numerator and denominator both approach \(0\) or \(\infty\). In our case, let \(f(x) = 1 - \cos x\) and \(g(x)=2x\). Now we compute their derivatives: $$ f'(x) = \frac{d}{dx} (1-\cos x) = \sin x $$ $$ g'(x) = \frac{d}{dx} (2x) = 2 $$ Now, we can find the limit of the ratio of the derivatives as \(x\) approaches \(0\): $$ \lim _{x \to 0} \frac{\sin x}{2} = \frac{1}{2} \lim _{x \to 0} \sin x $$ The limit of \(\sin x\) as \(x\) approaches \(0\) is \(0\), thus: $$ \frac{1}{2} \lim _{x \to 0} \sin x = \frac{1}{2} \cdot 0 = 0 $$
04

Determine the value of a for g(x) to be continuous

Using the result from Step 3, we know that \(\lim _{x \to 0} \frac{1-\cos x}{2x} = 0\). Now, for \(g(x)\) to be continuous at \(x=0\), we must have $$ g(0) = \lim _{x \to 0} \frac{1-\cos x}{2x} $$ Since g(0) is defined to be a, we obtain: $$ a = \lim _{x \to 0} \frac{1-\cos x}{2x} = 0 $$ To make \(g(x)\) continuous, it must be the case that \(g(0)=a=0\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

L'Hôpital's Rule
In calculus, L'Hôpital's rule is a popular technique for evaluating limits that result in indeterminate forms, such as \frac{0}{0}\ or \frac{\infty}{\infty}\. When faced with the challenge of determining the limit of a fraction where both the numerator and denominator approach zero or infinity, this rule becomes extremely handy.

The rule states that under certain conditions, the limit of a ratio of two functions can be found by taking the limit of the ratio of their derivatives. Specifically, it's expressed as:

\[\begin{equation}\lim_{{x \to a}} \frac{{f(x)}}{{g(x)}} = \lim_{{x \to a}} \frac{{f'(x)}}{{g'(x)}}\end{equation}\]
where both \( f(x) \) and \( g(x) \) must be differentiable and approach 0 or \( \infty \) as \( x \) approaches \( a \), and the limit of \( \frac{{f'(x)}}{{g'(x)}} \) exists or is \( \infty \).

If these conditions are met, L'Hôpital's rule greatly simplifies the original limit problem, turning something potentially complex into a more manageable calculation. Keep in mind, sometimes more than one application of the rule is necessary to resolve the indeterminate form.
Limits of Functions
The limit of a function is a fundamental concept in calculus, which describes the behavior of a function as its argument approaches a particular point. The notation \(\lim_{{x \to c}} f(x)\) is used to indicate the value that \(f(x)\) approaches as \(x\) approaches \(c\).

Limits help us understand the behavior of functions near points that might not be directly reachable, either because the function doesn't exist at that point, or because the point is at infinity. They serve as the foundation for defining continuity, derivatives, and integrals — the cornerstones of calculus.

In our context, the limit \(\lim_{{x \to 0}} \frac{{1 - \cos x}}{{2x}}\) is sought to determine where the function \(g(x)\) is continuous. If this limit exists and is equal to the value of the function at the point of interest, in this case, \(g(0)\), then we can say the function is continuous at that point.
Indeterminate Forms
When attempting to evaluate limits, mathematicians often encounter indeterminate forms — expressions whose limits cannot be determined solely from the limits of their individual parts. The most common indeterminate forms are \(0/0\), \(0\times\infty\), \(\infty - \infty\), \(\infty/\infty\), \(1^\infty\), \(\infty^0\), and \(0^0\).

In the step-by-step solution provided above, we tackle the indeterminate form \(0/0\) arising from \(\lim_{{x \to 0}} \frac{{1 - \cos x}}{{2x}}\). This kind specifies that both numerator and denominator approach zero independently, but we cannot conclude the behavior of the fraction without further analysis. Indeterminate forms challenge us to think deeper and often require the use of special techniques like L'Hôpital's rule or algebraic manipulation to find a limit that makes sense out of the ambiguity.
Derivatives
A derivative of a function is a measure of how a function's output value changes as the input changes. In formal terms, the derivative of the function \(f(x)\) with respect to \(x\) is defined as the limit of the difference quotient as the increment in \(x\) approaches zero:

\[\begin{equation}f'(x) = \lim_{{\Delta x \to 0}} \frac{{f(x + \Delta x) - f(x)}}{{\Delta x}}\end{equation}\]
Derivatives are fundamental in representing rates of change and finding slopes of tangent lines to curves. They play a central role in differential calculus and are used in a range of mathematical, scientific, and engineering fields.

In our exercise, we calculated the derivatives of \(1 - \cos x\) and \(2x\) in order to apply L'Hôpital's rule, which are \(\sin x\) and \(2\), respectively. The derivative of \(\cos x\) is \(-\sin x\), and for a constant multiplied by \(x\), the derivative is just the constant itself. Through these derivatives, we were able to resolve the indeterminate form and find the limit essential for determining the continuity of the function \(g(x)\).

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