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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo
Capitolo 7, Problema 7.5.71

Theory and Applications
L’Hôpital’s Rule does not help with the limits in Exercises 69–76. 
Try it—you just keep on cycling. Find the limits some other way.
71. lim (x → (π/2)⁻) sec x / tan x

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1
Identify the limit expression: \(\lim_{x \to (\pi/2)^-} \frac{\sec x}{\tan x}\).
Rewrite the trigonometric functions in terms of sine and cosine to simplify the expression: \(\sec x = \frac{1}{\cos x}\) and \(\tan x = \frac{\sin x}{\cos x}\). Substitute these into the limit to get \(\lim_{x \to (\pi/2)^-} \frac{\frac{1}{\cos x}}{\frac{\sin x}{\cos x}}\).
Simplify the complex fraction by multiplying numerator and denominator appropriately: \(\frac{\frac{1}{\cos x}}{\frac{\sin x}{\cos x}} = \frac{1}{\cos x} \times \frac{\cos x}{\sin x} = \frac{1}{\sin x}\).
Now the limit reduces to \(\lim_{x \to (\pi/2)^-} \frac{1}{\sin x}\). Consider the behavior of \(\sin x\) as \(x\) approaches \(\pi/2\) from the left side.
Since \(\sin x\) approaches 1 as \(x\) approaches \(\pi/2\), analyze the limit \(\frac{1}{\sin x}\) accordingly to determine the limit value.

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Behavior of Trigonometric Functions Near Specific Points

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Evaluating limits as x approaches π/2 from the left (denoted π/2⁻) requires considering the function's behavior just before π/2, which can differ from approaching from the right, impacting the limit's existence and value.
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