Problem 66 - Single Variable Calculus

Question

Show that $$ \lim _{x \rightarrow 0} \frac{x^{2} \sin \left(\frac{1}{x}\right)}{\sin x}=0 $$

Step-by-Step Solution

Answer

By rewriting the given expression as a product of two functions, \(f(x) = g(x) \cdot h(x)\), where \(g(x) = x^{2}\) and \(h(x) = \sin(\frac{1}{x})\cdot\frac{1}{\sin(x)}\), we find that \(\lim_{x \rightarrow 0} g(x) = 0\), while the limit of \(h(x)\) as x approaches 0 does not exist. However, the product of these limits still results in the limit of the given expression being 0. Thus, \(\lim _{x \rightarrow 0} \frac{x^{2} \sin \left(\frac{1}{x}\right)}{\sin x}\) = 0.

Step 1: Rewrite the expression as a product of two functions

Let's rewrite the given expression as \(f(x) = g(x) \cdot h(x)\), where \[g(x)=x^{2}\] and \[h(x) = \sin\left(\frac{1}{x}\right)\cdot\frac{1}{\sin(x)}\]

Step 2: Find limits of individual functions

The limit of the function g(x) as x approaches 0 is straightforward: \[\lim_{x \rightarrow 0} g(x) = \lim_{x \rightarrow 0} x^{2} = 0\] The limit of the function h(x) as x approaches 0 is more complicated and requires further examination.

Step 3: Analyze the function h(x)

Observe that the function \(h(x)\) can be bounded by using the fact that \(|\sin(u)|\le 1\). Thus, we have: \[|h(x)| = \left|\sin\left(\frac{1}{x}\right)\cdot\frac{1}{\sin(x)}\right| \le \left|\frac{1}{\sin(x)}\right|\] As x approaches 0, \(\sin(x)\) also approaches 0, and therefore, we get: \[\lim_{x \rightarrow 0} h(x) = \frac{1}{0}\] The limit of \(h(x)\) does not exist as x approaches 0.

Step 4: Apply limit laws

Since the limit of g(x) is 0 and the limit of h(x) does not exist, we can use the following limit law to find the limit of the product f(x): \[\lim_{x \rightarrow 0} f(x) = \lim_{x \rightarrow 0} g(x) \cdot h(x) = 0 \cdot \frac{1}{0} = 0\] Thus, the limit of the given expression as x approaches 0 is 0. The result is: \[\lim _{x \rightarrow 0} \frac{x^{2} \sin \left(\frac{1}{x}\right)}{\sin x} = 0\]

Key Concepts

Squeeze Theorem

The Squeeze Theorem is a very useful tool in calculus. It helps find limits of functions that are difficult to evaluate directly. The basic idea is to "squeeze" the function you are interested in between two other functions that have a known limit. If the upper and lower functions get close to the same value as x approaches a point, then the function trapped between them must also converge to the same limit.

Think of it like trapping a ball between your hands. If both hands get closer to each other, the ball has no choice but to be squeezed to the same point as well.

• For any function f(x), if you can find two functions, L(x) and U(x) such that L(x) ≤ f(x) ≤ U(x) for all x in some interval around a point,
• And if \( \lim_{x \to c} L(x) = \lim_{x \to c} U(x) = L \),
• Then \( \lim_{x \to c} f(x) = L \).

In the specific exercise you are given, we can potentially apply the Squeeze Theorem on the part \( \sin\left(\frac{1}{x}\right) \) by noting how the sine values oscillate between -1 and 1. If this part of the function is bounded, the enclosing functions will limit the expression to 0 as x approaches 0.

Trigonometric Limits

Trigonometric limits are frequently encountered when evaluating the limit of expressions involving trigonometric functions. They require understanding how these functions behave as their input approaches certain values. Generally, features like periodicity and boundedness play a crucial role.

In our exercise, we encounter \( \sin\left(\frac{1}{x}\right) \), which is part of a more complex function. Trigonometric functions like sine and cosine are periodic, meaning they repeat their values in regular intervals. This periodicity sometimes results in the non-existence of straightforward limits because it just oscillates wildly instead of settling to a particular value. However, their bounded nature (for instance, sine always stays between -1 and 1) can be useful.

In the limit problem presented, we consider \( \sin\left(\frac{1}{x}\right) \) oscillating as x approaches 0. Yet the bounded property of the sine function plays a role given that the overall limit aims to produce zero due to the multiplication with another component that exclusively approaches zero (namely \( x^2 \)). This highlights the advantageous property of trigonometric bounds in tackling tricky limits.

Limit Laws

Limit Laws are foundational in calculus. They provide a set of tools for breaking down complex limits into manageable parts. When you face a complicated limit problem, referring to the limit laws can save lots of time and effort.

Here are some important limit laws to remember:

• The Constant Rule: \( \lim_{x \to c} k = k \), where k is a constant.
• The Sum Rule: \( \lim_{x \to c} [f(x) + g(x)] = \lim_{x \to c} f(x) + \lim_{x \to c} g(x) \).
• The Product Rule: \( \lim_{x \to c} [f(x)g(x)] = \lim_{x \to c} f(x) \cdot \lim_{x \to c} g(x) \).
• The Quotient Rule, as long as \( \lim_{x \to c} g(x) eq 0 \): \( \lim_{x \to c} \frac{f(x)}{g(x)} = \frac{\lim_{x \to c} f(x)}{\lim_{x \to c} g(x)} \).

In our exercise, the limit law for products plays a significant role. Even though \( \lim_{x \to 0} h(x) \) does not exist independently, it does not affect the result due to \( \lim_{x \to 0} g(x) = 0 \). This exemplifies how leveraging limit laws can simplify a seemingly intricate limit into a simple resolution.