/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 47 $$\lim _{x \rightarrow \pi^{+}} ... [FREE SOLUTION] | 91影视

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Chapter 1: Problem 47

$$\lim _{x \rightarrow \pi^{+}} \frac{\sqrt{x}}{\csc x}$$

Short Answer

Expert verified
The limit as x approaches \(\pi^{+}\) of \(\frac{\sqrt{x}}{\csc x}\) is 0.

Step by step solution

01

Substitution

Try to substitute \(\pi^{+}\) into the function directly, you will find that the denominator \(\csc (\pi^{+})\) becomes undefined.
02

manipulate the given equation

Replace \(\csc x\) with \(\frac{1}{\sin x}\) because the secant function is the reciprocal of the sine function. Hence, the equation becomes \(\frac{\sqrt{x}}{\frac{1}{\sin x}}\) or \(\sqrt{x} \cdot \sin x \).
03

subbing the limit into the transformed function

Now, sub in the limit \(\pi^{+}\) into \(\sqrt{x} \cdot \sin x \). It would result in \(\sqrt{\pi^{+}} \cdot \sin (\pi^{+})\). Since \(\sin (\pi) \) is 0, and the limit from the right of \(\sqrt{x}\) at \(\pi^{+}\) is \(\sqrt{\pi}\), therefore the limit is \(\sqrt{\pi} \cdot 0\).

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Key Concepts

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

Substitution method
The substitution method is a common technique in calculus used to simplify the evaluation of limits. When faced with a limit problem, we start by substituting the point of interest directly into the function equation. This method allows us to check if the function simplifies or presents any indeterminate forms like 0/0 or p{undefined}{undefined}{}.
In the given exercise, the first step was to substitute a{href='https://en.wikipedia.org/wiki/Plus_sign_%28mathematics%29#Plus_sign_after_a_number'}{'+ '}{ a}{ . } right after ( a}{ right after adding it. This action helped us determine how the denominator behaves as it approaches i point it wasn't directly solvable due to the undefined nature of csc{ ( a}{ . br> br> To proceed, the function required a rewrite to eliminate the undefined term. This guarantees that the limit can be evaluated without error or ambiguity.
Trigonometric functions
Trigonometric functions play a crucial role in calculus, especially when dealing with periodic functions and limits. In this case, we worked with the cosecant function, rac{{1}}{ { . n order to simplify the original limit problem efficiently, we capitalized on the reciprocal identity: replacing csc{x}{ (x{|} rac{{1}{r>}} { c{ r> This transformation from cosecant to sine was pivotal in our solution, as sine is continuous at , helping eliminate the undefined behavior br> a{x}{c { , which undergoes straightforward evaluation.
Continuity and Discontinuity
Understanding continuity and discontinuity is vital when solving limits, as it tells us if a function behaves predictably around certain values. A function is continuous if it can be drawn without lifting the pen; otherwise, it's discontinuous.
In this exercise, the problematic point was ( r{as a}}{ confused with the concept of undefined limits. `` This understanding allowed us to correctly handle the sine function near { { in cosine or cosecant functions at specific values, leading to vertical asymptotes or undefined behavior Enhancing our approach, we evaluated the smoothed-out form ( 鈭毾� through a transformed, continuous perspective, enabling us to resolve the limit confidently through its continuity.

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