Problem 21 - Thomas Calculus

Question

Find the limits in Exercises \(11-22\) $$\lim _{h \rightarrow 0} \frac{3}{\sqrt{3 h+1}+1}$$

Step-by-Step Solution

Answer

The limit is \(\frac{3}{2}\).

Step 1: Understand the Problem

We need to find \(\lim_{h \rightarrow 0} \frac{3}{\sqrt{3h+1}+1}\). This involves evaluating the limit as \(h\) approaches zero.

Step 2: Rationalize the Denominator

To find the limit, eliminate the square root in the denominator by multiplying both the numerator and the denominator by the conjugate, \(\sqrt{3h+1} - 1\). This helps simplify the expression:\[\lim_{h \to 0} \frac{3}{\sqrt{3h+1}+1} \times \frac{\sqrt{3h+1} - 1}{\sqrt{3h+1} - 1}.\]

Step 3: Simplify the Expression

Expand the denominator using the difference of squares formula:\[(\sqrt{3h+1} + 1)(\sqrt{3h+1} - 1) = (3h+1) - 1 = 3h.\]This gives us the expression \[\lim_{h \to 0} \frac{3(\sqrt{3h+1} - 1)}{3h}.\]

Step 4: Cancel Out the Common Factor

Notice that \(3h\) in the denominator cancels with the 3 from the numerator:\[\lim_{h \to 0} \frac{\sqrt{3h+1} - 1}{h}.\]

Step 5: Apply the Limit Approach

Now, substitute \(0\) into the expression where \(h\) approaches 0, realizing that\[\lim_{h \to 0} \frac{\sqrt{3h+1} - 1}{h}\]can be evaluated using L'Hôpital's Rule because it's still in the \(\frac{0}{0}\) indeterminate form.

Step 6: Use L'Hôpital's Rule

According to L'Hôpital's Rule, differentiate the numerator and the denominator:- Derivative of the numerator \(\sqrt{3h+1} - 1\) is \(\frac{3}{2\sqrt{3h+1}}\).- Derivative of the denominator \(h\) is 1.Thus, the limit becomes:\[\lim_{h \to 0} \frac{3}{2\sqrt{3h+1}} = \frac{3}{2}\times 1 = \frac{3}{2}.\]

Step 7: Conclusion

The limit \(\lim_{h \to 0} \frac{3}{\sqrt{3h+1}+1}\) is \(\frac{3}{2}\).

Key Concepts

L'Hôpital's Rule

L'Hôpital's Rule is a powerful technique used in calculus to evaluate limits, especially when dealing with indeterminate forms like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). When direct substitution in a limit gives such an indeterminate form, L'Hôpital's Rule allows us to differentiate the numerator and the denominator separately, then find the limit of this new function.
This rule requires conditions to be met:

• The original limit must be in an indeterminate form.
• Both the numerator and denominator must be differentiable.
• The limit of their derivatives must exist.

Using L'Hôpital's Rule can considerably simplify the process of finding limits, such as the example provided where the indeterminate form \(\frac{0}{0}\) was resolved by differentiating the expression and calculating the resulting limit.

Rationalizing the Denominator

Rationalizing the denominator involves eliminating any irrational numbers, such as square roots, from the denominator of a fraction. This is useful in calculus for simplifying expressions to make limits easier to evaluate. Consider this process similar to clearing up conflicts at the base of the fraction, ensuring operations can proceed smoothly.
When a denominator involves a square root, like the one in \(\sqrt{3h+1}+1\), we multiply both the numerator and the denominator by the conjugate of the denominator.
Here's a quick recap on how it works:

• The conjugate of \(\sqrt{3h+1}+1\) is \(\sqrt{3h+1}-1\).
• Multiplying by the conjugate leverages the formula \((a+b)(a-b) = a^2 - b^2\) to clear square roots.

This trick transforms complex terms into something more manageable, setting the stage for further simplification steps like recognizing the difference of squares.

Difference of Squares

The difference of squares is a formula that allows the simplification of expressions that can be written as \(a^2 - b^2\). This formula is given by \(a^2 - b^2 = (a+b)(a-b)\). It's particularly handy in rationalizing denominators, as it transforms products involving square roots into simple expressions.
In the given exercise, once we rationalize the denominator with the conjugate \(\sqrt{3h+1} - 1\), we apply the difference of squares formula.

• This turns \((\sqrt{3h+1}+1)(\sqrt{3h+1}-1)\) into \((3h+1) - 1\).
• The simplification reduces the expression to just \(3h\), removing the square root and easing further simplification.

This step is essential for progressing towards a simpler form that is easier to work with for limit evaluation, eventually leading towards using L'Hôpital’s Rule effectively.