Problem 120
Question
$$ \lim _{x \rightarrow 0} \frac{\sin \left(x^{2}\right)}{x}\\{\text { Ans. } 0\\} $$
Step-by-Step Solution
Verified Answer
Applying the sine limit rule, we rewrite the function as \(f(x) = \frac{\sin(x^2)}{x^2} \cdot x\). The limit of \(\frac{\sin(x^2)}{x^2}\) as x approaches 0 is 1, and the limit of x as x approaches 0 is 0. Therefore, the limit of the original function \(f(x) = \frac{\sin(x^2)}{x}\) as x approaches 0 is \(1 \cdot 0 = 0\).
1Step 1: Rewrite the function using sine properties
We can rewrite the function as:
\(f(x) = \frac{\sin(x^2)}{x^2} \cdot x\)
2Step 2: Apply the limit rule for sine function
We know that:
\(\lim_{x \rightarrow 0} \frac{\sin(x)}{x} = 1\)
Since our function is rewritten as:
\(f(x) = \frac{\sin(x^2)}{x^2} \cdot x\)
We can apply the limit rule for sine function:
\(\lim_{x \rightarrow 0} \frac{\sin(x^2)}{x^2} = 1\)
3Step 3: Find the limit of the remaining factor
Now, we just need to find the limit of the remaining factor as x approaches 0:
\(\lim_{x \rightarrow 0} x = 0\)
4Step 4: Calculate the limit of original function
To find the limit of the original function, we can multiply the limits of the two factors we found previously:
\(\lim_{x \rightarrow 0} f(x) = \lim_{x \rightarrow 0} \left( \frac{\sin(x^2)}{x^2} \cdot x \right) = (\lim_{x \rightarrow 0} \frac{\sin(x^2)}{x^2}) \cdot (\lim_{x \rightarrow 0} x) = 1 \cdot 0 = 0\)
Thus, the limit of the given function as x approaches 0 is 0.
Key Concepts
Sine functionLimit rulesDifferential Calculus
Sine function
The sine function, denoted as \( \sin(x) \), is one of the fundamental functions in trigonometry. It describes a smooth repetitive oscillation that is essential in modeling cycles like sound and light waves. In calculus, the sine function is important because of its differentiability and periodic nature.
- The sine function repeats every \( 2\pi \) radians, meaning \( \sin(x + 2\pi) = \sin(x) \).- As we approach limits, especially near zero, sine behaves predictably, underpins several limit rules.
A common property used in calculus regarding sine is that \( \lim_{x \rightarrow 0} \frac{\sin(x)}{x} = 1 \). This property plays a crucial role when evaluating limits that involve small angles. Understanding the behavior of sine around small values is essential for various calculations in differential calculus.
- The sine function repeats every \( 2\pi \) radians, meaning \( \sin(x + 2\pi) = \sin(x) \).- As we approach limits, especially near zero, sine behaves predictably, underpins several limit rules.
A common property used in calculus regarding sine is that \( \lim_{x \rightarrow 0} \frac{\sin(x)}{x} = 1 \). This property plays a crucial role when evaluating limits that involve small angles. Understanding the behavior of sine around small values is essential for various calculations in differential calculus.
Limit rules
Limit rules are essential tools in calculus for determining the behavior of functions as they approach certain points. They help simplify complex mathematical expressions and find limits efficiently.
- The **product limit rule** states that \( \lim_{x \to a} [f(x) \cdot g(x)] = \lim_{x \to a} f(x) \cdot \lim_{x \to a} g(x) \), if both limits exist.- The **constant multiple rule** allows us to pull constants out of limit expressions: \( \lim_{x \to a} [c \cdot f(x)] = c \cdot \lim_{x \to a} f(x) \).
When solving problems like \( \lim_{x \rightarrow 0} \frac{\sin(x^2)}{x} \), we apply specific limit rules such as the trigonometric limit \( \lim_{x \rightarrow 0} \frac{\sin(x)}{x} = 1 \). These rules simplify analysis by breaking complex expressions into manageable parts.
- The **product limit rule** states that \( \lim_{x \to a} [f(x) \cdot g(x)] = \lim_{x \to a} f(x) \cdot \lim_{x \to a} g(x) \), if both limits exist.- The **constant multiple rule** allows us to pull constants out of limit expressions: \( \lim_{x \to a} [c \cdot f(x)] = c \cdot \lim_{x \to a} f(x) \).
When solving problems like \( \lim_{x \rightarrow 0} \frac{\sin(x^2)}{x} \), we apply specific limit rules such as the trigonometric limit \( \lim_{x \rightarrow 0} \frac{\sin(x)}{x} = 1 \). These rules simplify analysis by breaking complex expressions into manageable parts.
Differential Calculus
Differential calculus focuses on the concept of the derivative, which measures the rate of change of a function with respect to its variable. The derivative helps to understand how a function behaves as its input changes.
- A differential function is **continuous** and can be differentiated to find tangents, rates of change, and slopes.- The connection between derivatives and limits is central to calculus, as the derivative is defined as the limit of the difference quotient.
In problems involving limits and sine, such as the exercise \( \lim_{x \rightarrow 0} \frac{\sin(x^2)}{x} \), understanding derivatives allows us to predict how small changes in \( x \) will affect the function \( \sin(x^2) \). The skills gained in differential calculus apply directly to evaluating these types of limits and understanding the local behavior of functions.
- A differential function is **continuous** and can be differentiated to find tangents, rates of change, and slopes.- The connection between derivatives and limits is central to calculus, as the derivative is defined as the limit of the difference quotient.
In problems involving limits and sine, such as the exercise \( \lim_{x \rightarrow 0} \frac{\sin(x^2)}{x} \), understanding derivatives allows us to predict how small changes in \( x \) will affect the function \( \sin(x^2) \). The skills gained in differential calculus apply directly to evaluating these types of limits and understanding the local behavior of functions.
Other exercises in this chapter
Problem 118
$$ \lim _{x \rightarrow 1} \frac{\sin (\ln x)}{\ln x}\\{\text { Ans. } 1\\} $$
View solution Problem 119
$$ \lim _{x \rightarrow-\infty} \frac{\tan \left(e^{x}\right)}{e^{x}}\\{\text { Ans. } 1\\} $$
View solution Problem 121
$$ \lim _{x \rightarrow 0} \frac{\sin ^{2} 2 x}{x^{2}}\\{\text { Ans. } 4\\} $$
View solution Problem 122
$$ \lim _{x \rightarrow 0} \frac{1-\cos x}{x^{2}}\left\\{\text { Ans. } \frac{1}{2}\right. $$
View solution