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Chapter 5: Problem 21

If \(f(x)=e^{[\cot x]}\) where \([y]\) represents the greatest integer less than or equal to \(y\), then (a) \(\lim _{x \rightarrow \frac{\pi^{+}}{2}} f(x)=1\) (b) \(\lim _{x \rightarrow \frac{\pi^{+}}{2}} f(x)=\frac{1}{e}\) (c) \(\lim _{x \rightarrow \frac{x^{-}}{2}} f(x)=\frac{1}{e}\) (d) \(\lim _{x \rightarrow \frac{\pi^{-}}{2}} f(x)=1\)

Short Answer

Expert verified
The correct limits are: (a) and (c).

Step by step solution

01

Understanding the Function

We need to analyze the function \(f(x) = e^{[\cot x]}\), where \([y]\) is the greatest integer function, which returns the largest integer less than or equal to \(y\). Specifically, we need to determine how \(\cot x\) behaves as \(x\) approaches \(\pi/2\) from the right (\(\pi/2^+\)) and from the left (\(\pi/2^-\)).
02

Analyzing \(\cot x\) as \(x \to \frac{\pi}{2}^+\)

As \(x\) approaches \(\pi/2\) from the right, \(\cot x = \frac{1}{\tan x}\), and \(\tan x\) tends to negative infinity because \(\tan x\) rapidly decreases on this path. Consequently, \(\cot x\) tends to zero because \(\cot x\) is positive and decreases towards zero. Hence, \([\cot x] \rightarrow 0\).
03

Analyzing \(\cot x\) as \(x \to \frac{\pi}{2}^-\)

As \(x\) approaches \(\pi/2\) from the left, \(\tan x\) tends to positive infinity and \(\cot x = \frac{1}{\tan x}\) tends to zero from the negative side. Therefore, \([\cot x]\) still remains as zero because \(\cot x\) is close to but less than zero. Thus, \([\cot x] = -1\) before reaching zero.
04

Calculating \(\lim _{x \to \pi/2^+} f(x)\)

Now we calculate the limit: \[\lim_{x \to \pi/2^+} f(x) = \lim_{x \to \pi/2^+} e^{[\cot x]} = e^0 = 1.\] This confirms that the limit from the right hand side equals 1.
05

Calculating \(\lim _{x \to \pi/2^-} f(x)\)

For the left hand side of the limit: \[\lim_{x \to \pi/2^-} f(x) = \lim_{x \to \pi/2^-} e^{[\cot x]} = e^{-1} = \frac{1}{e}.\] This confirms that the limit from the left hand side equals \(\frac{1}{e}\).
06

Identifying the Correct Option

Considering the calculated limits: from the right \((\frac{\pi}{2}^+)\), \(\lim_{x \to \pi/2^+} f(x) = 1\); and from the left \((\frac{\pi}{2}^-)\), \(\lim_{x \to \pi/2^-} f(x) = \frac{1}{e}\). Thus, the correct options are (a) and (c).

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Key Concepts

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

Greatest Integer Function
The greatest integer function, also known as the floor function, is represented by the notation \([y]\). This function returns the largest integer that is less than or equal to any given real number \(y\). For example, \([3.7] = 3\) and \([鈭�1.2] = 鈭�2\).

In the provided exercise, the expression \(f(x) = e^{[\cot x]}\) utilizes the greatest integer function. Here, \([\cot x]\) gives us an integer value of \(\cot x\) nearest and lesser, when \(x\) approaches different sides of \(\frac{\pi}{2}\). This impacts how we calculate limits in this context since the behavior of \(\cot x\) changes near \(\frac{\pi}{2}\).
  • As \(x\) approaches \(\frac{\pi}{2}^+\), \(\cot x\) approaches zero from positive values, yielding \([\cot x] = 0\).
  • When \(x\) approaches \(\frac{\pi}{2}^-\), \(\cot x\) approaches zero from negative values, leading to \([\cot x] = -1\).
Trigonometric Limits
Trigonometric limits involve evaluating expressions where trigonometric functions like sine, cosine, and tangent are part of the function approaching a specific angle. Understanding how these limits behave is crucial in calculus.

In this case, we focus on the cotangent function, \(\cot x = \frac{1}{\tan x}\). As \(x\) moves closer and closer to \(\frac{\pi}{2}\), the behavior of \(\tan x\) affects \(\cot x\) drastically:
  • Approaching from the right, \(x \to \frac{\pi}{2}^+\), \(\tan x\) decreases sharply towards negative infinity, causing \(\cot x\) to trend towards zero from positive values.
  • From the left, \(x \to \frac{\pi}{2}^-\), \(\tan x\) increases towards positive infinity, making \(\cot x\) edge towards zero from the negative side.
These trigonometric limits help us interpret the function \(f(x)\) in terms of exponential expressions.
Exponential Functions
Exponential functions, such as \(e^x\), are critical in calculus due to their unique growth properties and the relationship to the natural logarithm. With \(f(x) = e^{[\cot x]}\), the value of \([\cot x]\) directly influences the behavior of the exponential function.

Analyzing the limits:
  • For \(\lim_{x \to \frac{\pi}{2}^+} f(x)\), we saw that \([\cot x] \rightarrow 0\), leading to \(e^0 = 1\).
  • For \(\lim_{x \to \frac{\pi}{2}^-} f(x)\), \([\cot x] = -1\), resulting in \(e^{-1} = \frac{1}{e}\).
The peculiar behaviors of these exponential functions at specific integer powers underscore the importance of understanding both the whole number powers and their corresponding exponential transformations. Each calculation reinforces how shifts in the base of \(\cot x\) affect \(f(x)\) dramatically, highlighting the role of exponential functions in defining limits.

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Most popular questions from this chapter

If \(\lim _{x \rightarrow a} f(x)=1\) and \(\lim _{x \rightarrow a} g(x)=\infty\), then \(\lim _{x \rightarrow a}\\{f(x)\\}^{g(x)}=e^{\lim (f(x)-1) \times g(x)}\) \(\lim _{x \rightarrow 0}\left(\frac{x-1+\cos x}{x}\right)^{\frac{1}{x}}\) is equal to (a) \(e^{1 / 2}\) (b) \(e^{-1 / 2}\) (c) \(e^{1}\) (d) \(\frac{1}{e}\)

If \(x_{1}=\sqrt{3}\) and \(x_{n+1}=\frac{x_{n}}{1+\sqrt{1+x_{n}^{2}}}, \forall n \in N\), then \(\lim _{x \rightarrow \infty} 2^{n} x_{n}\) equal to (a) \(\frac{3}{2 \pi}\) (b) \(\frac{2}{3 \pi}\) (c) \(\frac{2 \pi}{3}\) (d) \(\frac{3 \pi}{2}\)

Match the statements of Column I with values of Column II. Column I (A) \(\lim _{x \rightarrow \frac{\pi^{+}}{2}} \tan ^{-1}(\tan x)\) (B) \(\lim _{n \rightarrow \infty}\left[\sum_{r=1}^{n} \frac{1}{2^{r}}\right]([.]\) denotes the greatest integer function) (C) \(\lim _{x \rightarrow-} \sec ^{-1}\left(\frac{x}{x+1}\right)\) (D) \(\lim _{x \rightarrow \frac{\pi}{9}} \frac{\cos x}{(1-\sin x)^{2 / 3}}\) Column II (p) 0 (q) Doesn't exist (r) \(-\frac{\pi}{2}\) (s) \(\frac{\pi}{2}\)

The vlaue of \(\lim _{x \rightarrow 0} \frac{\sqrt{\sec ^{2} \frac{x}{2}-1}}{x}\) is (a) \(\frac{1}{2}\) (b) \(-\frac{1}{2}\) (c) 1 (d) doesn't exist

For \(n \in N\), let \(f_{n}(x)=\tan \frac{x}{2}(1+\sec x)(1+\sec 2 x)(1+\sec 4 x) \ldots\left(1+\sec 2^{n} x\right)\) Then, \(\lim _{x \rightarrow 0} \frac{f_{n}(x)}{2 x}\) is equal to (a) 0 (b) \(2^{n}\) (c) \(2^{n-1}\) (d) \(2^{n+1}\)
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