Problem 56 - Single Variable Calculus

Question

Find the limit, if it exists. \(\lim _{t \rightarrow 0} \frac{t}{\sqrt{2 t+1}-1}\)

Step-by-Step Solution

Answer

The limit of the given function as t approaches 0 is \(\boxed{1}\).

Step 1: Rationalize the denominator

To rationalize the denominator, we need to multiply both the numerator and the denominator by the conjugate of the denominator: \(\frac{\sqrt{2t + 1} + 1}{\sqrt{2t + 1} + 1}\)

Step 2: Simplify the expression

Now, we simply need to multiply the numerators and denominators with each other: \(\lim _{t \rightarrow 0} \frac{t(\sqrt{2t+1} + 1)}{(\sqrt{2t+1}-1)(\sqrt{2t+1}+1)}\) Next, apply the difference of squares formula to the denominator: \(\lim _{t \rightarrow 0} \frac{t(\sqrt{2t+1} + 1)}{2t}\)

Step 3: Simplify further and cancel out the common terms

Notice that we can cancel out the 't' term in both the numerator and the denominator: \(\lim _{t \rightarrow 0} \frac{\sqrt{2t+1} + 1}{2}\)

Step 4: Evaluate the limit

We can now easily evaluate the limit by substituting t with 0 in the expression: \(\frac{\sqrt{2(0)+1} + 1}{2}\) This simplifies to: \(\frac{\sqrt{1} + 1}{2}\) Finally, we find the limit to be: \(\frac{1 + 1}{2} = \boxed{1}\) So, the limit of the given function as t approaches 0 is 1.

Key Concepts

Rationalization

Rationalization is a useful technique when you're dealing with expressions that involve square roots in the denominator. In the context of limits, especially, it simplifies the expression and makes evaluating easier.
To rationalize, multiply both the numerator and the denominator by the conjugate of the denominator. The conjugate is the same expression as the denominator but with the opposite sign between terms.
For instance, if your denominator is \( \sqrt{2t+1} - 1 \), the conjugate would be \( \sqrt{2t+1} + 1 \). This is because the multiplication of a binomial and its conjugate results in a difference of squares, eliminating the square root in the process. This makes further simplifications much simpler.

• By removing the square roots, you obtain a polynomial or rational function, which is easier to simplify.
• This often helps in evaluating limits, as certain terms can cancel out, leaving expressions that are straightforward to solve.

When you multiply by the conjugate, ensure it's applied both in the numerator and denominator to avoid changing the fraction's value, effectively multiplying by 1.

Difference of Squares Formula

The Difference of Squares Formula is a key algebraic identity used to simplify expressions involving squared terms. It states that:
\[ (a-b)(a+b) = a^2 - b^2 \]
This formula transforms the product of two conjugates into the difference of their squares. In the context of limits and rationalization, it plays a crucial role.
When you multiply the expression \( (\sqrt{2t+1} - 1)(\sqrt{2t+1} + 1) \), the result is obtained as:

• \[ (\sqrt{2t+1})^2 - 1^2 = 2t + 1 - 1 = 2t \]

By applying this formula, you simplify the denominator from a complex square root-based expression to a simple polynomial \( 2t \). This makes it easier to factor, cancel common terms, and further simplifies the evaluation of limits.

Evaluating Limits

Evaluating limits involves finding the value that a function approaches as the variable gets arbitrarily close to a particular point.
Once you simplify an expression using techniques like rationalization and applying algebraic identities, evaluating limits becomes more straightforward.
For example, after removing any expression causing an indeterminate form, check if the expression is continuous at the point of evaluation.

• Substitute the limit's value into the simplified expression. If no indeterminate form persists, the evaluation is straightforward.
• Here, by substituting \( t = 0 \) in the simplified expression \( \frac{\sqrt{2t+1} + 1}{2} \), the radical simplifies to 1, as \( \sqrt{1} = 1 \).

This gives us \( \frac{1 + 1}{2} = 1 \), concluding that the limit is 1. The ability to manage expressions and seamlessly lead them to their limit aids in understanding both the process and result.