Problem 43 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find each of the right-hand and left-hand limits or state that they do not exist. $$\lim _{x \rightarrow 3^{+}} \frac{x-3}{\sqrt{x^{2}-9}}$$

Step-by-Step Solution

Answer

Right-hand limit is 0; left-hand limit does not exist.

Step 1: Analyze the Limit from the Right

The given expression is \( \lim _{x \rightarrow 3^{+}} \frac{x-3}{\sqrt{x^{2}-9}} \). As \( x \to 3^+ \), the denominator \( \sqrt{x^2 - 9} \) approaches \( \sqrt{3^2 - 9} = \sqrt{0} = 0 \). This makes the expression potentially undefined. We need to determine how the expression behaves as \( x \) approaches just above 3.

Step 2: Simplify the Expression for Limit from the Right

As \( x \approaches 3^+ \), write \( x = 3 + h \) where \( h \approaches 0^+ \). Substitute in the limit: \[ \lim_{h \to 0^+} \frac{(3+h)-3}{\sqrt{(3+h)^2-9}} = \lim_{h \to 0^+} \frac{h}{\sqrt{9 + 6h + h^2 - 9}} = \lim_{h \to 0^+} \frac{h}{\sqrt{6h + h^2}}. \]

Step 3: Evaluate the Behavior at Zero

In the expression \( \frac{h}{\sqrt{6h + h^2}} \), for \( h > 0 \), the square root \( \sqrt{6h + h^2} = \sqrt{h(6+h)} \). This can be rewritten as \( \sqrt{h} \sqrt{6+h} \). Thus, the expression becomes: \[ \lim_{h \to 0^+} \frac{h}{\sqrt{h} \sqrt{6+h}} = \lim_{h \to 0^+} \frac{\sqrt{h}}{\sqrt{6+h}}. \] As \( h \to 0^+ \), the expression simplifies to \( \frac{0}{\sqrt{6}} = 0 \).

Step 4: Conclude Right-Hand Limit

Therefore, the right-hand limit exists and \( \lim _{x \rightarrow 3^{+}} \frac{x-3}{\sqrt{x^{2}-9}} = 0 \).

Step 5: Analyze the Limit from the Left

Now, consider \( \lim _{x \rightarrow 3^{-}} \frac{x-3}{\sqrt{x^{2}-9}} \). For \( x \to 3^- \), we'll analyze the behavior similar to the previous steps.

Step 6: Simplify the Expression for Limit from the Left

Use \( x = 3 - h \) where \( h \to 0^+ \). Substitute in the limit: \[ \lim_{h \to 0^+} \frac{(3-h)-3}{\sqrt{(3-h)^2-9}} = \lim_{h \to 0^+} \frac{-h}{\sqrt{9 - 6h + h^2 - 9}} = \lim_{h \to 0^+} \frac{-h}{\sqrt{-6h + h^2}}. \]

Step 7: Evaluate the Behavior at Zero from the Left

For the expression \( \frac{-h}{\sqrt{-6h + h^2}} \), in \( h > 0 \), the square root can be expressed as \( \sqrt{h(-6+h)} \). This is not defined as real for small positive \( h \), indicating that the square root's argument becomes negative. Hence the limit does not exist.

Step 8: Conclude Left-Hand Limit

Since the expression \( \sqrt{h(-6+h)} \) is not defined in real numbers as \( h \to 0^+ \), the left-hand limit does not exist. Therefore, \( \lim _{x \rightarrow 3^{-}} \frac{x-3}{\sqrt{x^{2}-9}} \) does not exist.

Key Concepts

Right-hand limit

A right-hand limit refers to the behavior of a function as the input values approach a specific point from the right side, or from values greater than the target point. For example, with the expression \( \lim _{x \rightarrow 3^{+}} \frac{x-3}{\sqrt{x^{2}-9}} \), we are approaching 3 from values just slightly more than 3
(like 3.1 or 3.01). In the solution we've analyzed, we examine the behavior of the expression when \( x = 3 + h \) where \( h \) approaches zero from the positive side (\( h \to 0^+ \)). This gives us a new expression: \( \frac{h}{\sqrt{6h + h^2}} \). When we simplify this, the crucial observation is that both the numerator \( h \) and the square root part contain small positive quantities as \( h \to 0^+ \). This influences the limit to approach zero, resulting in a right-hand limit of 0. This means that when coming from the right towards 3, the given expression's value shrinks to zero.

Left-hand limit

Contrary to the right-hand limit, the left-hand limit focuses on the behavior of a function as the input values approach a specific point from the left, or from values less than the target point. If we consider \( \lim _{x \rightarrow 3^{-}} \frac{x-3}{\sqrt{x^{2}-9}} \), this means we are examining how the function behaves as \( x \) approaches 3 from values just below 3
(such as 2.9 or 2.99).To analyze this, we set \( x = 3 - h \) with \( h \to 0^+ \), meaning \( h \) approaches zero from the positive side. This results in the expression \( \frac{-h}{\sqrt{-6h + h^2}} \). However, the significant issue arises because \( -6h + h^2 \) turns negative as \( h \) gets close to zero from the positive side, leading to a negative value under the square root.Since the square root of a negative number is not defined within the real numbers, the expression cannot be evaluated, and thus, the left-hand limit does not exist. This means that as we approach 3 from the left, the expression does not settle into a predictable, finite value.

Limit does not exist

When we say a limit does not exist, it means that as we approach a specific input value, the function does not settle into a single predictable value. In this case with \( \lim _{x \rightarrow 3^{-}} \frac{x-3}{\sqrt{x^{2}-9}} \), the limit does not exist due to the square root of a negative number issue discussed previously.This is an example where mathematical operations cannot produce a real, finite result as you approach the point from one side. A limit might not exist because:

• The function heads off to infinity.


• The function oscillates between widely differing values.


• Technical mathematical reasons, like the square root of negative numbers here.

In essence, it’s crucial to conclude whether a limit exists with careful analysis, as simply substituting in values can often lead to incorrect conclusions when limits approach undefined or undefined-like behavior.