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Chapter 11: Problem 52

Find the following limits: (i) \(\lim _{x \rightarrow 0} \frac{x}{\tan x}.\) (ii) \(\lim _{x \rightarrow 0} \frac{\cos ^{2} x-1}{x^{2}}.\)

Short Answer

Expert verified
(i) 1;(ii) -1.

Step by step solution

01

Identify the limit expression (i)

Examine \(\lim _{x \rightarrow 0} \frac{x}{\tan x}.\)
02

Recall the trigonometric identity

Remember that \(\tan x = \frac{\sin x}{\cos x}.\)
03

Substitute the trigonometric identity into the limit expression (i)

Rewrite the limit as \(\lim _{x \rightarrow 0} \frac{x}{\frac{\sin x}{\cos x}} = \lim _{x \rightarrow 0} x \cdot \frac{\cos x}{\sin x}.\)
04

Simplify the limit expression (i)

Simplify the fraction to \(\frac{\cos x}{\sin x}.\)
05

Use the squeeze theorem for small angles (i)

Recall that \(\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1\). Finally, the limit becomes \(\lim_{x \rightarrow 0} \frac{x \cdot \cos x}{\sin x} = \cos(0) \cdot \lim_{x \rightarrow 0} \frac{x}{\sin x} = 1 \cdot 1 = 1. \)
06

Identify the limit expression (ii)

Examine \(\lim _{x \rightarrow 0} \frac{\cos ^2 x - 1}{x ^{2}}.\)
07

Use the trigonometric identity (ii)

Recall the identity \(\cos ^2 x - 1 = - \sin ^2 x.\)
08

Substitute the trigonometric identity into the limit expression (ii)

Rewrite the limit as \(\lim _{x \rightarrow 0} \frac{-\sin ^2 x}{x ^{2}}.\)
09

Simplify the limit expression (ii)

Rewrite the fraction as \(\lim _{x \rightarrow 0} -\left(\frac{\sin x}{x}\right)^2.\)
10

Apply the limit property (ii)

Since \(\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1,\) \(\left(\frac{\sin x}{x}\right)^2 = 1^2 = 1.\) Hence, the limit is \(\lim_{x \rightarrow 0} -1 = -1\).

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Key Concepts

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

Understanding Trigonometric Limits
Trigonometric limits are essential in calculus, especially when dealing with functions involving sine, cosine, and tangent. One common trigonometric limit is \( \lim_ {x \rightarrow 0} \frac{\text{sin} x}{x} = 1 \). This fundamental limit helps us solve more complex trigonometric limits. When evaluating limits like \( \lim_ {x \rightarrow 0} \frac{x}{\text{tan} x} \), it is crucial to rewrite the trigonometric function in a more manageable form using known identities, then simplify and apply these foundational limits. For small angles, approximations like \( \sin x \approx x \) are highly useful.
Applying Trigonometric Identities
Trigonometric identities simplify limit problems by transforming complex expressions. For example, in exercise (i), we use \( \tan x = \frac{\text{sin} x}{\text{cos} x} \). By substituting this identity into the limit \( \lim_ {x \rightarrow 0} \frac{x}{\text{tan} x} \, \) we get:

\ \lim_ {x \rightarrow 0} \frac{x}{\frac{\text{sin} x}{\text{cos} x}} = \lim_ {x \rightarrow 0} x \cdot\frac{\text{cos} x}{\text{sin} x} \.

This makes it easier to simplify and find the limit using fundamental limits. Using trigonometric identities such as \( \cos^2 x + \text{sin}^2 x = 1 \) and other Pythagorean identities can make the problem-solving process much smoother.
Squeeze Theorem and Its Applications
The Squeeze Theorem is a crucial tool for proving limits, especially when dealing with trigonometric functions at small values. The theorem states that if \(f(x) \leq g(x) \leq h(x)\) and \lim {x \rightarrow a} f(x) = \lim {x \rightarrow a} h(x) = L\, then \lim {x \rightarrow a} g(x) = L\. This is key to understanding why \lim_{x \rightarrow 0} \frac{\text{sin} x}{x} = 1 \. By 'squeezing' \frac{\text{sin} x}{x} \ between two other functions that converge to the same limit, we can firmly conclude the original limit. The limits in the exercise, such as \( \lim_ {x \rightarrow 0} \frac{x}{\text{tan} x}\), are easier to evaluate with this theorem.
Limit Properties and Their Importance
Understanding limit properties is essential for solving complex limit problems. Some key properties include:
  • Limits of sums: \ \lim_ {x \rightarrow a} [f(x) + g(x)] = \lim_ {x \rightarrow a} f(x) + \lim_ {x \rightarrow a} g(x) \.
  • Limits of products: \ \lim_ {x \rightarrow a} [f(x) \cdot g(x)] = \lim_ {x \rightarrow a} f(x) \cdot \lim_ {x \rightarrow a} g(x) \.
  • Constants: \ \lim_ {x \rightarrow a} c \cdot f(x) = c \cdot \lim_ {x \rightarrow a} f(x) \.
In problems like (ii) \(\ \lim_ {x \rightarrow 0} \frac{\text{cos}^2 x - 1}{x^2}\), we use these properties to decompose and simplify the limit. By converting \( \text{cos}^2 x - 1 \) into \( -\text{sin}^2 x \) and using known standard limits, we can systematically solve the problem. In this case, the limit simplifies to -1 by applying the property to the squared \( \frac{\text{sin} x}{x} \).

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Most popular questions from this chapter

(a) Suppose that \(f^{\prime}(x) > g^{\prime}(x)\) for all \(x,\) and that \(f(a)=g(a) .\) Show that \(f(x)> g(x)\) for \(x > a\) and \(f(x)< g(x)\) for \(x < a.\) (b) Show by an example that these conclusions do not follow without the hypothesis \(f(a)=g(a).\) (c) Suppose that \(f(a)=g(a),\) that \(f^{\prime}(x) \geq g^{\prime}(x)\) for all \(x,\) and that \(f^{\prime}\left(x_{0}\right)>\) \(g^{\prime}\left(x_{0}\right)\) for some \(x_{0}>a .\) Show that \(f(x)>g(x)\) for all \(x \geq x_{0}.\)

What is wrong with the following use of I'H么pital's Rule: $$\lim _{x \rightarrow 1} \frac{x^{3}+x-2}{x^{2}-3 x+2}=\lim _{x \rightarrow 1} \frac{3 x^{2}+1}{2 x-3}=\lim _{x \rightarrow 1} \frac{6 x}{2}=3.$$ (The limit is actually \(-4 .\))

Find the trapezoid of largest area that can be inscribed in a semicircle of radius \(a,\) with one base lying along the diameter.

(a) If \(a_{1} < \cdots< a_{n},\) find the minimum value of \(f(x)=\sum_{i=1}^{n}\left(x-a_{i}\right)^{2}.\) (b) Now find the minimum value of \(f(x)=\sum_{i=1}^{n}\left|x-a_{i}\right| .\) This is a problem where calculus won't help at all: on the intervals between the \(a_{i}\) 's the function \(f\) is linear, so that the minimum clearly occurs at one of the \(a_{i}\) and these are precisely the points where \(f\) is not differentiable. However, the answer is easy to find if you consider how \(f(x)\) changes as you pass from one such interval to another. (c) Let \(a>0 .\) Show that the maximum value of $$f(x)=\frac{1}{1+|x|}+\frac{1}{1+|x-a|}$$ is \((2+a) /(1+a) .\) (The derivative can be found on each of the intervals \((-\infty, 0),(0, a),\) and \((a, \infty)\) separately.)

A function \(f\) is increasing at \(a\) if there is some number \(\delta>0\) such $$f(x)>f(a) \text { if } a< x< a+\delta$$ and $$f(x)< f(a) \text { if } a-\delta < x < a$$ Notice that this does not mean that \(f\) is increasing in the interval \((a-\delta .\) \(a+\delta) ;\) for example, the function shown in Figure 34 is increasing at \(0,\) but is not an increasing function in any open interval containing 0. (a) Suppose that \(f\) is continuous on [0.1] and that \(f\) is increasing at \(a\) for every \(a\) in \([0.1] .\) Prove that \(f\) is increasing on \([0.1] .\) (First convince yourself that there is something to be proved.) Hint: For \(0< b <1\) prove that the minimum of \(f\) on \([b, 1]\) must be at \(b\). (b) Prove part (a) without the assumption that \(f\) is continuous, by considering for each \(b\) in [0.1] the set \(S_{b}=\\{x: f(y) \geq f(b) \text { for all } y \text { in }[b, x]\\}\) (This part of the problem is not necessary for the other parts.) Hint: Prove that \(S_{b}=\\{x: b \leq x \leq 1\\}\) by considering sup \(S_{b}.\) (c) If \(f\) is increasing at \(a\) and \(f\) is differentiable at \(a\), prove that \(f^{\prime}(a) \geq 0\) (this is easy). (d) II \(f^{\prime}(a)>0,\) prove that \(f\) is increasing at \(a\) (go right back to the definition of \(f^{\prime}(a)\).) (e) Use parts (a) and (d) to show, without using the Nean Value Theorem, that if \(f\) is continuous on \([0,1]\) and \(f^{\prime}(a)>0\) for all \(a\) in \([0.1],\) then \(f\) is increasing on \([0,1].\) (f) Suppose that \(f\) is continuous on [0,1] and \(f^{\prime}(a)=0\) for all \(a\) in (0,1) Apply part (e) to the function \(g(x)=f(x)+\varepsilon x\) to show that \(f(1)-\) \(f(0)>-\varepsilon .\) Similarly, show that \(f(1)-f(0)<\varepsilon\) by considering \(h(x)=\varepsilon x-f(x) .\) Conclude that \(f(0)=f(1).\) This particular proof that a function with zero derivative must be constant has many points in common with a proof of H. A. Schwarz, which may be the first rigorous proof ever given. Its discoverer, at least, seemed to think it was. See his exubcrant letter in reference [54] of the Suggested Reading.
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