Question

Other techniques Evaluate the following limits, where a and \(b\) are fixed real numbers. $$\lim _{h \rightarrow 0} \frac{\sqrt{16+h}-4}{h}$$

Short answer

Question: Evaluate the limit as h approaches 0 of the expression (sqrt(16+h)-4)/h. Answer: The limit equals -7/8.

Step-by-step solution

Identify indeterminate form

As we can see, when h approaches 0, we get: $$\lim _{h \rightarrow 0} \frac{\sqrt{16+h}-4}{h} = \frac{\sqrt{16+0}-4}{0} = \frac{0}{0}$$ Hence, we cannot directly find the limit by substitution, and we need to apply rationalization.

Rationalize the expression

To rationalize the expression, we will multiply the numerator and denominator by the conjugate of the numerator: $$\lim _{h\rightarrow 0} \frac{(\sqrt{16+h}-4)}{h} \times \frac{(\sqrt{16+h}+4)}{(\sqrt{16+h}+4)}$$

Simplify the rationalized expression

Now, multiply the numerators and denominators, and simplify the expression: $$\lim_{h \rightarrow 0} \frac{(16+h)-2 \cdot 4 \cdot \sqrt{16+h}+16}{h(\sqrt{16+h}+4)}$$ $$\lim_{h \rightarrow 0} \frac{h-8\sqrt{16+h}}{h(\sqrt{16+h}+4)}$$

Cancel out the h term

As we can see, the h term in the numerator can be cancelled out by the h term in the denominator: $$\lim_{h \rightarrow 0} \frac{1-8\frac{\sqrt{16+h}}{h}}{\sqrt{16+h}+4}$$

Evaluate the limit

Now we can directly substitute h=0 into the expression and find the limit: $$\lim_{h \rightarrow 0} \frac{1-8\frac{\sqrt{16+0}}{0}}{\sqrt{16+0}+4} = \frac{1-8(1)}{4+4}$$

Simplifying the expression

Finally, simplify the expression: $$\frac{1-8}{8} = \frac{-7}{8}$$ Thus, the limit is equal to: $$\lim _{h \rightarrow 0} \frac{\sqrt{16+h}-4}{h} = \frac{-7}{8}$$

Key concepts

Indeterminate forms

When solving limits, you might encounter situations where directly substituting the limit value into an expression results in an undefined form, like \( \frac{0}{0} \). This is known as an indeterminate form. It doesn't provide any conclusive information about the limit itself. Instead, it serves as a signal that tells us to try other techniques, such as algebraic manipulation, to evaluate the limit correctly. In our original problem, as \( h \) approaches 0, the expression \( \frac{\sqrt{16+h}-4}{h} \) becomes \( \frac{0}{0} \). This indicates an indeterminate form, and it means we cannot simply plug in \( h = 0 \) to find the solution. This requires us to attempt other methods to further analyze the expression, such as rationalization.

Rationalization

Rationalization is an algebraic technique used to eliminate radicals (like square roots) from the numerator or denominator of a fraction. This is particularly useful in limit problems where direct substitution results in an indeterminate form, like \( \frac{0}{0} \). To rationalize an expression, you multiply the numerator and denominator by the conjugate of the expression containing the radical. The conjugate is obtained by changing the sign between two terms. For example, the conjugate of \( \sqrt{16+h}-4 \) is \( \sqrt{16+h}+4 \). By multiplying both the numerator and the denominator by this conjugate, we are able to transform and simplify the expression, making it easier to evaluate the limit. In the given exercise, this step was crucial in moving past the indeterminate form.

Conjugate multiplication

Conjugate multiplication is the process of multiplying an expression by its conjugate, which effectively removes any radicals that might cause difficulties in calculation. The principle behind this technique is based on the difference of squares formula: \( (a-b)(a+b) = a^2 - b^2 \). This helps us to eliminate the square root in our expression by turning it into a simple polynomial form. For example, in the expression \( \sqrt{16+h} - 4 \), we multiply by \( \sqrt{16+h} + 4 \), resulting in \( (16+h) - 16 \), which simplifies to \( h \). This clever use of algebra lets us simplify complex expressions into ones that are easier to handle while evaluating limits. This step smooths the process of simplifying the expression further.

Simplifying expressions

Simplifying expressions is a fundamental skill in algebra that involves reducing expressions to their simplest form. This often involves canceling common factors, expanding products, and reducing fractions. In limit problems, simplification can turn an indeterminate form into something that is clearly defined and easy to work with. After rationalizing the expression \( \frac{\sqrt{16+h}-4}{h} \), and multiplying by the conjugate, we ended up with an expression that had \( h \) in both the numerator and the denominator. Canceling the \( h \) terms is a critical step to simplifying further and allows us to substitute \( h = 0 \) without resulting in division by zero. Simplification converts our tough initial expression into a straightforward one: \( \frac{1-8(1)}{4+4} \), which evaluates to \( \frac{-7}{8} \), giving us the final limit result.