Question

Linearization Show that the approximation of tan \(x\) by its linearization at the origin must improve as \(x \rightarrow 0\) by showing that $$\lim _{x \rightarrow 0} \frac{\tan x}{x}=1$$

Short answer

Yes, the approximation of tan \(x\) by its linearization at the origin improves as \(x \rightarrow 0\) as it has been shown that \(\lim _{x \rightarrow 0} \frac{\tan x}{x}=1\).

Step-by-step solution

Recall the limit of sin x/x

An important fact to remember is that when \(x \rightarrow 0\), \(\lim _{x \rightarrow 0} \frac{\sin x}{x}=1\). This is a well known result in calculus.

Express tan x as sin x / cos x

The function \(\tan x\) can be expressed as \(\frac{\sin x}{\cos x}\). So, we rewrite \(\tan x\) in this way which enables us to better manipulate the function and apply the limit. Thus, \(\frac{\tan x}{x}\) is the same as \(\frac{\sin x}{x \cdot \cos x}\).

Divide the function by x and apply limit

Next we divide the modified function by \(x\). We now have \(\lim _{x \rightarrow 0} \frac{\sin x / x}{\cos x}\). Since we know that \(\lim _{x \rightarrow 0} \frac{\sin x}{x}=1\), we can replace the \(\frac{\sin x}{x}\) -part in the fraction with 1. This yields \(\lim _{x \rightarrow 0} \frac{1}{\cos x}\).

Evaluate limit

Now, as \(x \rightarrow 0\), \(\cos(x)\) equals 1. So, \(\lim _{x \rightarrow 0} \frac{1}{\cos x} =1\). This completes the proof that the linear approximation of tan x improves as x approaches 0.

Key concepts

Limits in Calculus

Understanding the concept of limits is a cornerstone in calculus. It's a way to describe the behavior of a function as it approaches a certain point. Limits can be at the core of solving complex problems related to rates of change and finding the slope of curves.

In the context of linear approximation, limits help us confirm the accuracy of the approximation near a point. For example, in our exercise, the limit \( \text{lim}_{x\rightarrow 0} \frac{\tan(x)}{x} \) is evaluated to determine how closely \( \tan(x) \) can be approximated by a straight line when \( x \) is close to 0. By establishing that this limit is 1, we show that the linear approximation of \( \tan(x) \) by its tangent line at the origin (which is the x-axis in this case) becomes more accurate as \( x \) approaches 0.

Trigonometric Functions

Trigonometric functions are the bread and butter of many mathematical problems, especially when dealing with circles and waves. In our exercise, we're dealing with the tangent function, denoted as \( \tan(x) \) which is one of the primary trigonometric functions.

The tangent function is the ratio of the sides opposite and adjacent to the angle in a right triangle. Algebraically, it's expressed as \( \tan(x) = \frac{\sin(x)}{\cos(x)} \) when \( x \) is the angle. This identity allows us to simplify complex expressions and solve problems that involve angles and their relationships with triangles. As \( x \) approaches 0, the sine and cosine functions approach their values at 0, which are 0 and 1 respectively, making \( \tan(x) \) approach 0 as well. This behavior is critical when assessing the accuracy of the linear approximation of the \( \tan \) function.

Continuity and Differentiability

Continuity and differentiability are two concepts that tell us a lot about functions and their graphs. A function is continuous at a point if there is no break, jump, or hole at that point on the graph. Differentiability, on the other hand, means that a function has a derivative at that point, implying it has a tangent line that touches the curve without crossing it at that very point.

For a function to be differentiable, it must be continuous, but the converse is not always true. The linear approximation relies on both these characteristics. In our exercise, showing that \( \lim_{x\rightarrow 0} \frac{\tan(x)}{x} = 1 \) implies that \( \tan(x) \) is both continuous and differentiable at 0. As a result of these properties, the linear function that approximates \( \tan(x) \) at 0 is a good representation of \( \tan(x) \) when values of \( x \) are close to 0.