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Chapter 6: Problem 12

Try to use L'Hospital's Rule to find the limit of \(\frac{\tan x}{\sec x}\) as \(x \rightarrow(\pi / 2)-\). Then evaluate directly by changing to sines and cosines.

Short Answer

Expert verified
The limit of the \(\frac{\tan x}{\sec x}\) as \(x \rightarrow(\pi / 2)-\) is 1.

Step by step solution

01

Simplify the limit using trigonometric identities

The \(\tan{x} = \frac{\sin{x}}{\cos{x}}\) and \(\sec{x} = \frac{1}{\cos{x}}\). So, the limit function \(\frac{\tan x}{\sec x}\) can be written as \({\frac{\sin{x}}{\cos{x}}} \cdot \cos{x}\) which simplifies to \(\sin{x}\). The limit is now \(x \rightarrow(\pi / 2)-\) of \(\sin{x}\)
02

Evaluate the Limit

The \(\sin{x}\) function always lies between -1 and 1, inclusive. The limit as \(x \rightarrow(\pi / 2)-\) of \(\sin{x}\) is equal to 1. Hence, the limit of the function \(\frac{\tan x}{\sec x}\) as \(x \rightarrow(\pi / 2)-\) is 1.
03

Evaluate using L'Hopital's Rule

L'Hopital's rule states that the limit as \(x \rightarrow a\) of the ratio of two functions, both of which are either 0 or 鈭�, is equal to the limit as \(x \rightarrow a\) of the ratios of their derivatives. However, in this case, L'Hopital's rule doesn't need to be applied as we don't have the indeterminate forms 0/0 or 鈭�/鈭�. So, the first method itself holds for this problem.

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Let \(g(x):=\left|x^{3}\right|\) for \(x \in \mathbb{R}\). Find \(g^{\prime}(x)\) and \(g^{\prime \prime}(x)\) for \(x \in \mathbb{R}\), and \(g^{\prime \prime \prime}(x)\) for \(x \neq 0\). Show that \(g^{\prime \prime \prime}(0)\) does not exist.

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