Question

Find the indicated limit. Make sure that you have an indeterminate form before you apply l'Hôpital's Rule. $$ \lim _{t \rightarrow 1} \frac{\sqrt{t}-t^{2}}{\ln t} $$

Short answer

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

Step-by-step solution

Identify Indeterminate Form

First, substitute the value of the limit, which is 1, into the expression. This gives us \( \frac{\sqrt{1} - 1^2}{\ln 1} = \frac{1 - 1}{0} = \frac{0}{0} \). This is an indeterminate form, which allows us to apply l'Hôpital's Rule.

Apply l'Hôpital's Rule

Since we have confirmed an indeterminate form of \( \frac{0}{0} \), we apply l'Hôpital's Rule. This involves taking the derivative of the numerator and the derivative of the denominator and then finding the limit again.

Differentiate the Numerator

For the numerator \( \sqrt{t} - t^2 \), take the derivative: \( \frac{d}{dt}(\sqrt{t}) = \frac{1}{2\sqrt{t}} \) and \( \frac{d}{dt}(t^2) = 2t \). So the derivative of the numerator is \( \frac{1}{2\sqrt{t}} - 2t \).

Differentiate the Denominator

For the denominator \( \ln t \), the derivative is \( \frac{d}{dt}(\ln t) = \frac{1}{t} \).

Evaluate the Limit Again

Now evaluate the limit of the new expression: \( \lim_{t \to 1} \frac{\frac{1}{2\sqrt{t}} - 2t}{\frac{1}{t}} \). Substitute \( t = 1 \) into the new expression, giving: \( \frac{\frac{1}{2\cdot1} - 2\cdot1}{1} = \frac{\frac{1}{2} - 2}{1} = -\frac{3}{2} \). Thus, the limit is -\( \frac{3}{2} \).

Key concepts

Indeterminate Forms

In calculus, limits sometimes result in expressions like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\) when a function approaches a particular value. These are called "indeterminate forms." They don't directly tell us what the limit is, and further analysis is required. Indeterminate forms are a clue that something interesting happens at those points, and special techniques, like l'Hôpital's Rule, need to be applied. When faced with an indeterminate form, it's essential not to jump to conclusions but rather to apply mathematical methods that provide clarity. It's crucial to initially substitute the limit value to confirm if you indeed have an indeterminate form. As seen in our exercise, substituting \( t = 1 \) into \( \frac{\sqrt{t} - t^2}{\ln t} \) gave us \( \frac{0}{0} \), confirming we had an indeterminate form and paving the way to use l'Hôpital's Rule effectively.

Differentiation

Differentiation is the process of finding a derivative, which represents the rate at which a function is changing at any given point. In the context of limits, differentiation helps simplify expressions that initially present indeterminate forms. To tackle an indeterminate form using l'Hôpital's Rule, we differentiate the numerator and the denominator separately. This step often simplifies the expression such that its limit becomes determinable. For instance, in the numerator \( \sqrt{t} - t^2 \), applying differentiation gives us \( \frac{1}{2\sqrt{t}} - 2t \). Similarly, differentiating the denominator \( \ln t \) results in \( \frac{1}{t} \). With these derivatives, a new, often non-indeterminate expression is formed, which helps us evaluate the limit more straightforwardly.

Limits in Calculus

Limits are foundational in understanding calculus. They describe the behavior of a function as its input approaches a particular point. The concept of limits helps us understand the continuity of functions, the computation of derivatives, and the evaluation of integrals.When using l'Hôpital's Rule, it's all about assessing the limit of the derivatives when an indeterminate form appears. After differentiating both parts of the problematic expression, we re-evaluate the limit. In our exercise, after applying l'Hôpital’s Rule and reevaluating the limit as \( t \) approaches 1, we found that it simplifies to -\( \frac{3}{2} \). This process highlights how limits allow us to resolve complexities that arise in calculus.