Problem 32 - Calculus Early Transcendentals

Question

Evaluate the following limits. \(\lim _{h \rightarrow 0} \frac{3}{\sqrt{16+3 h}+4}\)

Step-by-Step Solution

Answer

Short Answer: The limit of the given expression, \(\lim _{h \rightarrow 0} \frac{3}{\sqrt{16+3 h}+4}\), is \(\frac{3}{8}\).

Step 1: Write down the given expression

We are given the following expression to evaluate the limit: \(\lim _{h \rightarrow 0} \frac{3}{\sqrt{16+3 h}+4}\)

Step 2: Rationalize the denominator

To simplify the expression, we can rationalize the denominator by multiplying both the numerator and denominator by the conjugate of the denominator, which is \(\sqrt{16+3h}-4\). This will eliminate the square root in the denominator so that we have an easier expression to work with: $$\lim _{h \rightarrow 0} \frac{3}{\sqrt{16+3 h}+4} * \frac{\sqrt{16+3h}-4}{\sqrt{16+3h}-4}$$

Step 3: Simplify the expression

Now, we multiply the numerators and denominators together and simplify: $$ \lim _{h \rightarrow 0} \frac{3(\sqrt{16+3 h}-4)}{(\sqrt{16+3 h}+4)(\sqrt{16+3 h}-4)} $$ In the denominator, we have a difference of squares which is \((a+b)(a-b) = a^2 - b^2\). Applying this property, we get: $$ \lim _{h \rightarrow 0} \frac{3(\sqrt{16+3 h}-4)}{(16+3 h) - 16} $$ Now, we can simplify further by canceling out the constant term in the denominator: $$ \lim _{h \rightarrow 0} \frac{3(\sqrt{16+3 h}-4)}{3 h} $$ Finally, cancel the common factor of 3 in the numerator and denominator: $$ \lim _{h \rightarrow 0} \frac{\sqrt{16+3 h}-4}{h} $$

Step 4: Evaluate the limit

Now that we have simplified the expression, we can plug in \(h=0\) to evaluate the limit: $$ \lim _{h \rightarrow 0} \frac{\sqrt{16+3(0)}-4}{0} = \frac{\sqrt{16}-4}{0} = \frac{4-4}{0} = \frac{0}{0} $$ However, this is an indeterminate form, so we need to use the L'Hopital's rule to find the limit. Applying L'Hopital's rule, we need to differentiate the numerator and the denominator of the expression with respect to \(h\): $$ \lim _{h \rightarrow 0} \frac{\frac{d}{dh}(\sqrt{16+3 h}-4)}{\frac{d}{dh}(h)} $$ Differentiating the expressions, $$ \frac{d}{dh}(\sqrt{16+3 h}-4) = \frac{3}{2\sqrt{16+3h}} \\ \frac{d}{dh}(h) = 1 $$ Thus, $$ \lim _{h \rightarrow 0} \frac{\frac{3}{2\sqrt{16+3h}}}{1} $$ Now, when \(h = 0\), we get, $$ \lim _{h \rightarrow 0} = \frac{3}{2\sqrt{16}} = \frac{3}{8} $$ Hence, the limit of the given expression is \(\frac{3}{8}\).

Key Concepts

Rationalizing the Denominator

To simplify expressions that involve roots in the denominator, we can utilize a technique known as rationalizing the denominator. This method primarily deals with eliminating the radical expressions from the denominator of a fraction to make it easier to handle. By multiplying both the numerator and the denominator by the conjugate of the denominator, we achieve this. The conjugate of a binomial expression like \( \sqrt{a} + b \) is \( \sqrt{a} - b \).

• This operation turns the denominator into a difference of squares, allowing us to eliminate the square root.
• It simplifies calculations by avoiding division by a binomial with a radical.

Overall, this technique helps us progress towards solving the expression, ensuring a more straightforward computation.

Indeterminate Forms

When calculating limits, we sometimes encounter expressions that are not immediately solvable, typically as a result of forms such as \( \frac{0}{0} \), known as indeterminate forms. In these cases, the solution is not directly visible.

• Indeterminate forms require additional manipulation or rules before we can find the limit.
• Initial simplification, such as canceling common factors or using other algebraic methods, can often help.

Once an indeterminate form is recognized, further techniques like L'Hôpital’s Rule can be applied to find the limit's actual value.

L'Hopital's Rule

L'Hopital's Rule is a practical method for solving limits that result in indeterminate forms. This elegant and straightforward rule states that if a limit of the form \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) is encountered, we can find the limit by differentiating the numerator and denominator separately, and then finding the limit of their quotient.The steps for applying L'Hôpital’s Rule include:

• Ensuring that the function is continuous and differentiable near the point of interest.
• Taking the derivative of both the numerator and the denominator.
• Re-evaluating the limit of the derivative terms.

Using this rule provides a powerful approach for handling tricky limits that don't resolve through basic algebraic manipulation, leading us directly to the solution.