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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.3, Problem 17
PROBLEM:
Evaluate \displaystyle \lim\limits_{x\to 0}\left(\displaystyle \frac{\sin\left(x\right)\sin\left(2x\right)}{1-\cos\left(x\right)}\right).
Direct substitution of x=0 gives the indeterminate form \frac{0}{0}. Therefore, we should apply trigonometric identities.
We know that \sin\left(2x\right)=2\sin\left(x\right)\cos\left(x\right), so we can rewrite the original function as
\begin{align*} \displaystyle \lim\limits_{x\to 0}\left(\displaystyle \frac{\sin\left(x\right)\sin\left(2x\right)}{1-\cos\left(x\right)}\right) & =\lim\limits_{x\to 0}\left(\frac{\sin\left(x\right)\cdot 2\left(\sin\left(x\right) \cos\left(x\right)\right)}{1-\cos\left(x\right)}\right)\\ \\ & =\displaystyle 2\cdot \lim\limits_{x\to 0}\left(\frac{\sin^2\left(x\right)\cos\left(x\right)}{1-\cos\left(x\right)}\right)\\ \end{align*}
We also know the Pythagorean identity \sin^2\left(x\right)=1-\cos^2\left(x\right). So,
\begin{align*} \displaystyle \lim\limits_{x\to 0}\left(\displaystyle \frac{\sin\left(x\right)\sin\left(2x\right)}{1-\cos\left(x\right)}\right) & =2\cdot \lim\limits_{x\to 0}\left(\frac{\left(1-\cos^2\left(x\right)\right)\cos\left(x\right)}{1-\cos\left(x\right)}\right)\\ \\ & =2\cdot \lim\limits_{x\to 0}\left(\frac{\left(1+\cos\left(x\right)\right)\left(1-\cos\left(x\right)\right)\cos\left(x\right)}{1-\cos\left(x\right)}\right) \\ \\ & =2\cdot \lim\limits_{x\to 0}\left(\left(1+\cos\left(x\right)\right)\cos\left(x\right)\right) \\ \\ & = 2\cdot \left(\left(1+\cos\left(0\right)\right)\cos\left(0\right)\right) \\ \\ & =2\cdot \left(\left(1+1\right)\cdot 1\right) \\ \\ & =4 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
Differential and Integral Calculus by Feliciano and Uy, Exercise 1.3, Problem 16
PROBLEM:
Evaluate \displaystyle \lim\limits_{x\to 0}\left(\frac{1-\cos^2\left(x\right)}{1+\cos\left(x\right)}\right)
This problem can be solved using a direct substitution of x=0. That is
\begin{align*} \lim\limits_{x\to 0}\left(\frac{1-\cos^2 x }{1+\cos x }\right) & =\frac{1-\cos^2\left(0\right)}{1+\cos\left(0\right)} \\ \\ & =\frac{1-1}{1+1}\\ \\ & = 0 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
Differential and Integral Calculus by Feliciano and Uy, Exercise 1.3, Problem 15
PROBLEM:
Evaluate \displaystyle \lim_{x\to 0}\left(\frac{\sin^3x}{\sin x-\tan x}\right).
SOLUTION:
A straight substitution of \displaystyle x=\frac{\pi }{4} leads to the indeterminate form \displaystyle \frac{0}{0} which is meaningless.
Therefore, to evaluate the limit of the given function, we proceed as follows
\begin{align*} \displaystyle \lim _{x\to 0}\left(\frac{\sin^3x}{\sin x-\tan x}\right) & =\lim _{x\to 0}\left(\frac{\sin^3x}{\sin x-\frac{\sin x}{\cos x}}\right) \\ \\ & =\lim _{x\to 0}\left(\frac{\sin^3 x}{\frac{\sin x \cos x - \sin x}{\cos x}}\right) \\ \\ & = \lim _{x\to 0}\left(\frac{\sin^3 x \cos x }{\sin x \cos x - \sin x}\right) \\ \\ &=\lim _{x\to \:0}\left(\frac{\sin^3 x \cos x}{\left(\sin x \right)\left(\cos x-1\right)}\right) \\ \\ & =\lim _{x\to 0}\left(\frac{\sin^2 x \cos x }{\left(\cos x -1\right)}\right) \\ \\ & =\lim _{x\to 0}\left(\frac{\left(1-\cos^2 x \right) \cdot\cos x }{-\left(1-\cos x \right)}\right) \\ \\ & =\lim\limits_{x\to 0}\left(\frac{\left(1+\cos x \right) \left(1-\cos x \right) \cdot\cos x }{-\left(1-\cos x \right)}\right) \\ \\ & =\lim _{x\to 0}\left(\frac{\left(1+\cos x \right) \cdot \cos\left(x\right)}{-1}\right) \\ \\ & =-1\cdot \lim _{x\to 0}\left(\left(1+\cos x\right)\cdot\cos x \right) \\ \\ & =-1\cdot \left(1+\cos 0 \right)\cdot \cos 0 \\ \\ & =-1\cdot \left(1+1\right)\cdot 1 \\ \\ & =-2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
Differential and Integral Calculus by Feliciano and Uy, Exercise 1.3, Problem 14
PROBLEM:
Evaluate \displaystyle \lim\limits_{x\to \frac{\pi }{4}}\left(\frac{\tan\:2x}{\sec\:2x}\right).
SOLUTION:
A straight substitution of x=\frac{\pi }{4} leads to the indeterminate form \frac{0}{0} which is meaningless.
Therefore, to evaluate the limit of the given function, we proceed as follows
\begin{align*} \displaystyle \lim\limits_{x\to \:\frac{\pi \:}{4}}\left(\frac{\tan\:2x}{\sec\:2x}\right) & =\lim\limits_{x\to \frac{\pi }{4}}\left(\frac{\frac{\sin\:2x}{\cos\:2x}}{\frac{1}{\cos\:2x}}\right) \\ \\ & =\lim\limits_{x\to \:\frac{\pi \:}{4}}\left(\sin\:2x\right) \\ \\ & =\sin\left(2\cdot \frac{\pi }{4}\right) \\ \\ & =\sin\frac{\pi }{2} \\ \\ & =1 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
Differential and Integral Calculus by Feliciano and Uy, Exercise 1.3, Problem 13
PROBLEM:
Evaluate \displaystyle \lim\limits_{x\to 3}\left(\frac{\sqrt{x^2-9}}{x-3}\right).
A straight substitution of x=3 leads to the indeterminate form \frac{0}{0} which is meaningless.
Therefore, to evaluate the limit of the given function, we proceed as follows
\begin{align*} \lim\limits_{x\to \:3}\left(\frac{\sqrt{x^2-9}}{x-3}\right) & =\lim\limits_{x\to 3}\left(\frac{\sqrt{x^2-9}}{x-3}\cdot \frac{\sqrt{x^2-9}}{\sqrt{x^2-9}}\right) \\ \\ & =\lim\limits_{x\to 3}\left(\frac{\left(x+3\right)\left(x-3\right)}{\left(x-3\right)\sqrt{x^2-9}}\right) \\ \\ & =\lim\limits_{x\to 3}\left(\frac{x^2-9}{\left(x-3\right)\sqrt{x^2-9}}\right) \\ \\ & =\lim _{x\to 3}\left(\frac{x+3}{\sqrt{x^2-9}}\right) \\ \\ & =\frac{3+3}{\sqrt{3^2-9}} \\ \\ & =\frac{6}{0} \\ \\ & =\infty \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
Since the function’s limit is different from the left to its limits from the right, the limit does not exist.
Chapter 1: Limits
Exercise 1.1: Functional Notation
Exercise 1.2: Theorems on Limits
Exercise 1.3: Indeterminate Forms
Exercise 1.4: Limit at Infinity
Exercise 1.5: Continuity
Problem 1
Problem 2
Problem 3
Problem 4
Problem 5
Problem 6
Exercise 1.6: Asymptotes
Problem 1
Problem 2
Problem 3
Problem 4
Problem 5
Problem 6
Problem 7
Problem 8
Problem 9
Problem 10
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