Problem 69 - Precalculus with Limits

Question

In Exercises 69 and 70, state which limit can be evaluated by using direct substitution. Then evaluate or approximate each limit. (a) \\[ \lim_{x \to 0}\ x^2 \sin x^2 \\] (b) \\[ \lim_{x \to 0}\ \dfrac{\sin\ x^2}{x^2} \\]

Step-by-Step Solution

Answer

The limit of the function \(x^2 \sin x^2\) as x approaches 0 is 0. The limit of the function \(\dfrac{\sin\ x^2}{x^2}\) as x approaches 0 is 1.

Step 1: Evaluate the limit of first function.

The first function is \(x^2 \sin x^2\). As the limit is considered at x=0, which is in the domain of the function, direct substitution can be used. So, substituting x=0 into the function, we get \(0^2 \sin 0^2\), which simplifies to 0. This is our limit value.

Step 2: Evaluate the limit of the second function.

The second function is \(\dfrac{\sin\ x^2}{x^2}\). Direct substitution results in the form 0/0, which is an indeterminate form. We can apply L'Hopital's rule here, which states that the limit of a ratio of two functions is equal to the limit of the ratios of their derivatives, if the limit is in indeterminate form. Taking the derivative of both the numerator and the denominator, we get \(\dfrac{2x \cos x^2}{2x}\). This simplifies to \(\cos x^2\). Now we can apply direct substitution, substituting x=0, we get \(\cos 0^2\), which simplifies to 1. This is our limit value.

Step 3: State the results.

The limit of the first function \(x^2 \sin x^2\) as x approaches 0 is 0 and the limit of the second function \(\dfrac{\sin\ x^2}{x^2}\) as x approaches 0 is 1.

Key Concepts

Direct Substitution

When you encounter a limit problem, the first step is often to evaluate whether direct substitution is applicable. This method involves simply substituting the value that x approaches into the function. It's a straightforward technique that works well if the function is continuous at that point and the substitution doesn’t result in an indeterminate form.

For instance, consider the limit \[ \lim_{x \to 0} x^2 \sin x^2. \]Since substituting 0 directly into x does not lead to any undefined value or indeterminate form, you can proceed with it directly:

• Substitute x = 0 to get \(0^2 \sin 0^2\).
• This clearly simplifies to 0, as \(\sin 0 = 0\).

If direct substitution works smoothly without creating undefined expressions, it’s always the preferred choice due to its simplicity.

L'Hopital's Rule

Sometimes, direct substitution leads to forms like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). These are indeterminate forms, and this is where L'Hopital's Rule comes in handy. This rule can be applied if the limit results in one of these indeterminate forms. It states that if the limit of a function results in 0/0 or \(\infty/\infty\), the limit can be evaluated by finding the derivatives of the numerator and denominator separately and then taking their limit.

Consider the limit \[ \lim_{x \to 0} \dfrac{\sin x^2}{x^2}. \]Direct substitution here provides the indeterminate form \(\frac{0}{0}\).

• Apply L'Hopital's Rule: take the derivative of the numerator, \(\sin x^2\), which gives \(2x \cos x^2\), and the derivative of the denominator, \(x^2\), which gives \(2x\).
• The limit then becomes \[ \lim_{x \to 0} \dfrac{2x \cos x^2}{2x}, \]
• which simplifies to \(\lim_{x \to 0} \cos x^2 = \cos 0 = 1\).

This shows how L'Hopital's Rule helps resolve an indeterminate form to give us the limit.

Indeterminate Forms

Indeterminate forms like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\) arise when direct substitution doesn't yield a clear result. If after substituting, both the numerator and denominator approach zero, or both approach infinity, the expression can't be directly evaluated.

These forms indicate potential complexity in the behavior of the function around the point of interest. They call for techniques beyond direct substitution to evaluate the limit accurately. Some approaches include:

• L'Hopital’s Rule, as explained earlier.
• Algebraic manipulation, such as factoring or rationalizing, to simplify the expression before substitution.

Understanding and identifying these forms is crucial as it tells you when additional calculation strategies are needed. These strategies help peel back the layers of mathematical complexity to find the simple truth underneath - the limit.