Calculate the Limit of f(x) as x approaches 0

The problem asks us to find the limit of the function \(f(x) = \frac{x^2+x+|x|}{x}\) as \(x\) approaches 0. To evaluate this limit, we need to consider the behavior of the function as \(x\) gets close to 0 from both the positive and negative sides. This is because the absolute value function, \(|x|\), is defined differently for positive and negative values of \(x\).

Analyzing the Function f(x) near x = 0

The function is given by \(f(x) = \frac{x^2+x+|x|}{x}\). Since \(|x|\) changes its definition at \(x=0\), we must evaluate the left-hand limit (LHL) and the right-hand limit (RHL) separately.

Right-Hand Limit (RHL) as x → 0⁺

When \(x\) approaches 0 from the positive side, denoted as \(x \rightarrow 0^+\), it means \(x > 0\). For \(x > 0\), the absolute value \(|x|\) is equal to \(x\). Substituting \(|x|=x\) into the function \(f(x)\), we get:

\(f(x) = \frac{x^2+x+x}{x}\)

\(f(x) = \frac{x^2+2x}{x}\)

For \(x \neq 0\), we can factor out \(x\) from the numerator and cancel it with the denominator:

\(f(x) = \frac{x(x+2)}{x}\)

\(f(x) = x+2\)

Now, we can find the limit as \(x \rightarrow 0^+\):

\(\displaystyle\lim_{x \rightarrow 0^+} f(x) = \lim_{x \rightarrow 0^+} (x+2)\)

Substituting \(x=0\), we get:

\(\displaystyle\lim_{x \rightarrow 0^+} f(x) = 0+2 = 2\)

So, the right-hand limit is 2.

Left-Hand Limit (LHL) as x → 0⁻

When \(x\) approaches 0 from the negative side, denoted as \(x \rightarrow 0^-\), it means \(x < 0\). For \(x < 0\), the absolute value \(|x|\) is equal to \(-x\). Substituting \(|x|=-x\) into the function \(f(x)\), we get:

\(f(x) = \frac{x^2+x+(-x)}{x}\)

\(f(x) = \frac{x^2+x-x}{x}\)

\(f(x) = \frac{x^2}{x}\)

For \(x \neq 0\), we can factor out \(x\) from the numerator and cancel it with the denominator:

\(f(x) = \frac{x \cdot x}{x}\)

\(f(x) = x\)

Now, we can find the limit as \(x \rightarrow 0^-\):

\(\displaystyle\lim_{x \rightarrow 0^-} f(x) = \lim_{x \rightarrow 0^-} x\)

Substituting \(x=0\), we get:

\(\displaystyle\lim_{x \rightarrow 0^-} f(x) = 0\)

So, the left-hand limit is 0.

Comparing Left-Hand and Right-Hand Limits

For the overall limit \(\displaystyle\lim_{x \rightarrow 0} f(x)\) to exist, the left-hand limit and the right-hand limit must be equal. In this case, we found:

• Right-Hand Limit (\(\displaystyle\lim_{x \rightarrow 0^+} f(x)\)) = 2
• Left-Hand Limit (\(\displaystyle\lim_{x \rightarrow 0^-} f(x)\)) = 0

Since the LHL (\(0\)) is not equal to the RHL (\(2\)), the limit \(\displaystyle\lim_{x \rightarrow 0} f(x)\) does not exist.

Conclusion on the Limit of f(x)

Because the limit from the left side of 0 is different from the limit from the right side of 0 for the function \(f(x)\), the overall limit as \(x\) approaches 0 does not exist.

Therefore, \(\displaystyle\lim_{x \rightarrow 0} f(x)\) does not exist.

Revision Table: Limits and Absolute Value

Additional Information: When Limits Do Not Exist

A limit \(\displaystyle\lim_{x \rightarrow a} f(x)\) does not exist if any of the following conditions are met:

• The left-hand limit (\(\displaystyle\lim_{x \rightarrow a^-} f(x)\)) exists but is not equal to the right-hand limit (\(\displaystyle\lim_{x \rightarrow a^+} f(x)\)). This was the case in our example with \(a=0\).
• The function approaches infinity or negative infinity as \(x\) approaches \(a\).
• The function oscillates infinitely as \(x\) approaches \(a\).

Understanding LHL and RHL is crucial when dealing with functions that have different definitions or behaviors on either side of the limit point, such as functions involving absolute values, piecewise functions, or step functions.