Question

Other techniques Evaluate the following limits, where a and \(b\) are fixed real numbers. $$\lim _{t \rightarrow a} \frac{\sqrt{3 t+1}-\sqrt{3 a+1}}{t-a}$$

Short answer

Answer: \(\frac{3}{2\sqrt{3a+1}}\)

Step-by-step solution

Rationalize the numerator

To eliminate the square roots from the numerator, we will multiply and divide the fraction by the conjugate of the numerator, which is \((\sqrt{3t+1} + \sqrt{3a+1})\). This will result in the following expression: $$\frac{\sqrt{3 t+1}-\sqrt{3 a+1}}{t-a} \cdot \frac{\sqrt{3 t+1}+\sqrt{3 a+1}}{\sqrt{3 t+1}+\sqrt{3 a+1}}$$

Multiply and Simplify

Now, we can multiply the numerators and denominators, resulting in the following expression: $$\frac{(3t+1)-(3a+1)}{(t-a)(\sqrt{3 t+1}+\sqrt{3 a+1})}$$ Next, we can simplify the numerator, which will leave us with: $$\frac{3t-3a}{(t-a)(\sqrt{3 t+1}+\sqrt{3 a+1})}$$

Simplify further

We can now factor out the 3 in the numerator, resulting in the following expression: $$\frac{3(t-a)}{(t-a)(\sqrt{3 t+1}+\sqrt{3 a+1})}$$ Now, we can cancel out the (t-a) terms in the numerator and denominator: $$\frac{3}{\sqrt{3 t+1}+\sqrt{3 a+1}}$$

Evaluate the limit

As t approaches a, the expression becomes: $$\lim _{t \rightarrow a} \frac{3}{\sqrt{3 t+1}+\sqrt{3 a+1}}$$ Substitute a for t: $$\frac{3}{\sqrt{3 a+1}+\sqrt{3 a+1}}$$ Finally, combine the terms in the denominator: $$\frac{3}{2\sqrt{3 a+1}}$$ The result is the limit of the given function as t approaches a: $$\lim _{t \rightarrow a} \frac{\sqrt{3 t+1}-\sqrt{3 a+1}}{t-a} = \frac{3}{2\sqrt{3 a+1}}$$

Key concepts

Rationalization

Rationalization is a powerful technique employed to eliminate square roots or other radicals from the numerator or denominator of a fraction. This is particularly useful in calculus for simplifying expressions and evaluating limits. In the context of the given exercise, we have a limit expression where the numerator contains square roots:

\( \sqrt{3 t+1} - \sqrt{3 a+1} \). To rationalize such an expression, we look for its conjugate. The conjugate of a binomial like \( a-b \) is \( a+b \). Here, the conjugate is \( \sqrt{3 t+1} + \sqrt{3 a+1} \).

When we multiply the expression by its conjugate, rather than affecting the overall value of the limit, we turn the radicals into a difference of squares. This step is essential as it simplifies the expression to a form where further algebraic manipulation, such as canceling terms, becomes possible. Rationalization helps reveal the underlying structure of the expression, making it easier to handle and evaluate the limit.

Indeterminate Forms

In calculus, indeterminate forms are expressions where direct substitution to evaluate a limit does not lead to a clear answer. An example of such a form is \( \frac{0}{0} \), often encountered in limit problems. These forms are important indicators that further algebraic manipulation or different techniques are needed to analyze the behavior of the function as it approaches a certain point.

In the problem provided, substituting \( t = a \) in the expression \( \frac{\sqrt{3 t+1} - \sqrt{3 a+1}}{t-a} \) directly leads to this indeterminate form. This signals that straightforward evaluation won't work, as we would essentially divide zero by zero.

Recognizing an indeterminate form is crucial since it guides us to employ suitable techniques, such as factorization, L'Hôpital's Rule, or the conjugate method as applied in this solution, to resolve the indeterminacy and find the true value of the limit.

Conjugate Method

The conjugate method is a specific technique often used in tandem with rationalization to simplify expressions involving square roots or other binomials. By utilizing the conjugate, we transform the initial expression into a more manageable form. The procedure is centered around the identity \((a-b)(a+b) = a^2 - b^2\), called the difference of squares.

For the exercise at hand, the conjugate method was critical. We multiplied both the numerator and the denominator of\[ \frac{\sqrt{3t+1} - \sqrt{3a+1}}{t-a} \]by \( \sqrt{3t+1} + \sqrt{3a+1} \). This effectively cleared the radicals from the numerator, yielding the form\[ \frac{(3t+1) - (3a+1)}{(t-a)(\sqrt{3t+1} + \sqrt{3a+1})} \]which simplifies to \[ \frac{3(t - a)}{(t-a)(\sqrt{3t+1}+\sqrt{3a+1})} \].

After this simplification, the conjugate method allows for the cancellation of \( t-a \) from both the numerator and the denominator, directly resolving the indeterminate form encountered earlier. This method is indispensable in limit problems involving radicals because it provides a clear pathway from an initially indeterminate form to a defined and calculable expression.