Understanding the Limit Problem

The question asks us to evaluate the limit of the function \(\rm \dfrac{\sin x \log (1-x)}{x^2}\) as \(x\) approaches \(0\). This is a common type of problem in calculus involving indeterminate forms.

First, let's check the form of the expression as \(x \rightarrow 0\):

• As \(x \rightarrow 0\), \(\sin x \rightarrow \sin 0 = 0\).
• As \(x \rightarrow 0\), \(\log (1-x) \rightarrow \log (1-0) = \log 1 = 0\).
• As \(x \rightarrow 0\), \(x^2 \rightarrow 0^2 = 0\).

So, the limit is of the indeterminate form \(\dfrac{0 \cdot 0}{0} = \dfrac{0}{0}\). We can use techniques like L'Hopital's Rule or standard limit formulas/series expansions to evaluate this limit.

Method for Evaluating the Limit

A straightforward way to evaluate this limit is by using known standard limits. Recall the following standard limits:

• \(\displaystyle\lim_{x\rightarrow 0} \dfrac{\sin x}{x} = 1\)
• \(\displaystyle\lim_{y\rightarrow 0} \dfrac{\log (1+y)}{y} = 1\)

We can manipulate the given expression to utilize these standard forms.

Step-by-Step Limit Calculation

Let's rewrite the given limit expression:

\[ \displaystyle\lim_{x\rightarrow 0} \dfrac{\sin x \log (1-x)}{x^2} \]

We can split the denominator \(x^2\) into two parts, \(x \cdot x\), and group terms to match the standard limit forms:

\[ \displaystyle\lim_{x\rightarrow 0} \left( \dfrac{\sin x}{x} \right) \cdot \left( \dfrac{\log (1-x)}{x} \right) \]

Now, we can evaluate the limit of each part separately, provided each limit exists:

Part 1: \(\displaystyle\lim_{x\rightarrow 0} \dfrac{\sin x}{x}\)

This is a standard limit:

\[ \displaystyle\lim_{x\rightarrow 0} \dfrac{\sin x}{x} = 1 \]

Part 2: \(\displaystyle\lim_{x\rightarrow 0} \dfrac{\log (1-x)}{x}\)

This limit is related to the standard limit \(\displaystyle\lim_{y\rightarrow 0} \dfrac{\log (1+y)}{y} = 1\). Let \(y = -x\). As \(x \rightarrow 0\), \(y \rightarrow 0\).

Then, \(\log(1-x) = \log(1+y)\) and \(x = -y\).

The limit becomes:

\[ \displaystyle\lim_{y\rightarrow 0} \dfrac{\log (1+y)}{-y} = \lim_{y\rightarrow 0} (-1) \cdot \dfrac{\log (1+y)}{y} \]

Using the standard limit \(\displaystyle\lim_{y\rightarrow 0} \dfrac{\log (1+y)}{y} = 1\):

\[ \lim_{y\rightarrow 0} (-1) \cdot \dfrac{\log (1+y)}{y} = (-1) \cdot 1 = -1 \]

So, \(\displaystyle\lim_{x\rightarrow 0} \dfrac{\log (1-x)}{x} = -1\).

Combining the parts:

Now, multiply the limits of the two parts:

\[ \displaystyle\lim_{x\rightarrow 0} \dfrac{\sin x \log (1-x)}{x^2} = \left( \lim_{x\rightarrow 0} \dfrac{\sin x}{x} \right) \cdot \left( \lim_{x\rightarrow 0} \dfrac{\log (1-x)}{x} \right) \]

\[ = (1) \cdot (-1) = -1 \]

Alternatively, using Taylor series expansions around \(x=0\):

• \(\sin x = x - \dfrac{x^3}{3!} + O(x^5) = x + O(x^3)\)
• \(\log(1-x) = -x - \dfrac{x^2}{2} - \dfrac{x^3}{3} + O(x^4) = -x + O(x^2)\)

Multiply the leading terms:

\[ \sin x \log(1-x) \approx (x)(-x) = -x^2 \text{ for small } x \]

Then, the limit is approximately:

\[ \displaystyle\lim_{x\rightarrow 0} \dfrac{-x^2}{x^2} = -1 \]

Using higher order terms:

\[ \sin x \log(1-x) = (x + O(x^3))(-x + O(x^2)) = -x^2 + O(x^3) \]

So, \(\dfrac{\sin x \log(1-x)}{x^2} = \dfrac{-x^2 + O(x^3)}{x^2} = -1 + O(x)\). As \(x \rightarrow 0\), this approaches \(-1\).

Result

The limit \(\rm \displaystyle\lim_{x\rightarrow 0} \dfrac{\sin x \log (1-x)}{x^2}\) is equal to \(-1\).

Revision Table: Key Standard Limits

Additional Information on Indeterminate Forms and Limits

When evaluating limits, certain forms like \(\dfrac{0}{0}\), \(\dfrac{\infty}{\infty}\), \(0 \cdot \infty\), \(\infty - \infty\), \(0^0\), \(1^\infty\), and \(\infty^0\) are called indeterminate forms. These forms do not immediately tell us the value of the limit, and further analysis is required.

Common techniques for evaluating limits with indeterminate forms include:

• Factoring and Cancelling: Useful for rational functions.
• Multiplying by Conjugates: Often used for expressions involving square roots.
• Using Standard Limits: Applying known limits like \(\displaystyle\lim_{x\rightarrow 0} \dfrac{\sin x}{x} = 1\) or \(\displaystyle\lim_{x\rightarrow 0} \dfrac{\log(1+x)}{x} = 1\). This was the primary method used in the solution above.
• L'Hopital's Rule: If \(\displaystyle\lim_{x\rightarrow c} \dfrac{f(x)}{g(x)}\) is of the form \(\dfrac{0}{0}\) or \(\dfrac{\infty}{\infty}\), then \(\displaystyle\lim_{x\rightarrow c} \dfrac{f(x)}{g(x)} = \lim_{x\rightarrow c} \dfrac{f'(x)}{g'(x)}\), provided the latter limit exists. While applicable here, it would involve differentiating products, which can be more complex than using standard limits directly.
• Taylor Series Expansions: Expanding functions like \(\sin x\), \(\log(1-x)\), \(e^x\), etc., into power series around the limit point (here, \(x=0\)) can simplify the expression and allow for easy evaluation of the limit by cancelling terms.

Choosing the right technique depends on the structure of the function. Recognizing standard limits often provides the most efficient path for certain problems.