91Ó°ÊÓ

Use L'Hóspital's Rule, where appropriate, to evaluate the limits a. \(\lim _{t \rightarrow \infty} \frac{e^{2 t}}{t} \quad\) b. \(\quad \lim _{t \rightarrow \infty} \frac{e^{t}}{\sqrt{t}} \quad\) c. \(\quad \lim _{t \rightarrow \infty} \frac{2 t^{2}+1}{5 t^{2}+2}\) d. \(\lim _{t \rightarrow \infty} \frac{\ln t}{e^{t}} \quad\) e. \(\quad \lim _{t \rightarrow 0^{+}} \frac{\ln t}{1 / t}\) f. \(\quad \lim _{t \rightarrow 0^{+}} \frac{\sin t}{t}\) \(\mathrm{g}\). \(\lim _{t \rightarrow \infty} \frac{\ln \sqrt{t}}{\sqrt{t}} \quad\) h. \(\quad \lim _{t \rightarrow 0^{+}} \frac{\sin 2 t}{\sin 3 t} \quad\) i. \(\quad \lim _{t \rightarrow \infty} \frac{\ln t}{\sqrt{t}}\) j. \(\quad \lim _{t \rightarrow \infty} \frac{3^{t}-1}{2^{t}-1}\) k. \(\lim _{t \rightarrow 0^{+}} \frac{t}{\ln (1+t)} \quad\) l. \(\quad \lim _{t \rightarrow 0+} \frac{\tan t}{\sqrt{t}}\)

Step by step solution

01

Evaluate Limit a

The given limit is \(\lim _{t \rightarrow \infty} \frac{e^{2 t}}{t}\). As \(t \to \infty\), both the numerator \(e^{2t}\) and the denominator \(t\) tend to infinity, so we have an \(\frac{\infty}{\infty}\) form. We can use L'Hôpital's Rule. Differentiate the numerator and denominator: \(\frac{d}{dt}e^{2t} = 2e^{2t}\) and \(\frac{d}{dt}t = 1\). Applying L'Hôpital's Rule gives the new limit \(\lim_{t\rightarrow\infty}\frac{2e^{2t}}{1} = \lim_{t\rightarrow\infty}2e^{2t} = \infty\).

02

Evaluate Limit b

The given limit is \(\lim _{t \rightarrow \infty} \frac{e^{t}}{\sqrt{t}}\). Both the numerator \(e^t\) and the denominator \(\sqrt{t}\) tend to infinity, forming an \(\frac{\infty}{\infty}\) indeterminate form. Differentiate the numerator and denominator: \(\frac{d}{dt}e^t = e^t\) and \(\frac{d}{dt}\sqrt{t} = \frac{1}{2\sqrt{t}}\). Applying L'Hôpital's Rule gives the new limit \(\lim_{t\rightarrow\infty} \frac{e^t}{\frac{1}{2\sqrt{t}}} = \lim_{t\rightarrow\infty} 2e^t\sqrt{t}\), which tends to \(\infty\).

03

Evaluate Limit c

The given limit is \(\lim _{t \rightarrow \infty} \frac{2t^2+1}{5t^2+2}\). Both the numerator and the denominator tend to infinity as \(t\to\infty\). Divide every term by \(t^2\): \(\lim_{t\rightarrow\infty}\frac{2 + \frac{1}{t^2}}{5 + \frac{2}{t^2}} = \frac{2}{5}\) as each term containing \(\frac{1}{t^2}\) approaches \(0\). So, the limit is \(\frac{2}{5}\).

04

Evaluate Limit d

The given limit is \(\lim _{t \rightarrow \infty} \frac{\ln t}{e^t}\). As \(t \to \infty\), both the numerator \(\ln t\) and denominator \(e^t\) go to infinity, but \(e^t\) grows much faster. Using L'Hôpital's Rule by differentiating gives \(\frac{d}{dt}\ln t = \frac{1}{t}\) and \(\frac{d}{dt}e^t = e^t\). This gives the new limit \(\lim_{t\rightarrow\infty} \frac{1/t}{e^t} = 0\).

05

Evaluate Limit e

The given limit is \(\lim _{t \rightarrow 0^{+}} \frac{\ln t}{1 / t}\). Rewriting it, \(\lim_{t\rightarrow 0^{+}} t \ln t\). As \(t \to 0^+\), \(t \ln t\) approaches \(0\). Hence, the limit is \(0\).

06

Evaluate Limit f

The given limit is \(\lim _{t \rightarrow 0^{+}} \frac{\sin t}{t}\). This is a standard limit that equals 1, as \(t \rightarrow 0^{+}\). So, the limit is \(1\).

07

Evaluate Limit g

The given limit is \(\lim _{t \rightarrow \infty} \frac{\ln \sqrt{t}}{\sqrt{t}}\). Rewrite it as \(\lim_{t\rightarrow\infty}\frac{\ln t^{1/2}}{\sqrt{t}} = \lim_{t\rightarrow\infty}\frac{\frac{1}{2}\ln t}{\sqrt{t}}\). This limit forms \(\frac{\infty}{\infty}\), so using L'Hôpital's gives \(\lim_{t\rightarrow\infty}\frac{\frac{1}{2t}}{\frac{1}{2\sqrt{t}}} = \lim_{t\rightarrow\infty}\frac{1}{t^{3/2}} = 0\).

08

Evaluate Limit h

The given limit is \(\lim _{t \rightarrow 0^{+}} \frac{\sin 2t}{\sin 3t}\). Using \(\sin x \approx x\) for small \(x\), rewrite the limit as \(\lim_{t\rightarrow0^{+}} \frac{2t}{3t} = \frac{2}{3}\).

09

Evaluate Limit i

The given limit is \(\lim _{t \rightarrow \infty} \frac{\ln t}{\sqrt{t}}\). Both tend to infinity, so apply L'Hôpital's Rule: differentiate to get \(\frac{1/t}{1/(2\sqrt{t})}=\frac{2}{\sqrt{t}}\), and the limit as \(t \to \infty\) becomes \(0\).

10

Evaluate Limit j

The given limit is \(\lim _{t \rightarrow \infty} \frac{3^{t}-1}{2^{t}-1}\). This forms \(\frac{\infty}{\infty}\). Differentiate both terms: \(\frac{d}{dt}(3^t) = 3^t\ln 3\) and \(\frac{d}{dt}(2^t) = 2^t\ln 2\). Now the limit becomes \(\lim_{t\rightarrow \infty} \frac{3^t\ln 3}{2^t\ln 2} = \lim_{t\rightarrow\infty} (\frac{3}{2})^t \frac{\ln 3}{\ln 2}\) which tends to \(\infty\) due to exponential growth.

11

Evaluate Limit k

The given limit is \(\lim _{t \rightarrow 0^{+}} \frac{t}{\ln (1+t)}\). This forms \(\frac{0}{0}\), so use L'Hôpital's Rule. Differentiate to get \(\frac{1}{\frac{1}{1+t}}=1+t\), and at \(t \to 0^+\) the limit becomes \(1\).

12

Evaluate Limit l

The given limit is \(\lim _{t \rightarrow 0+} \frac{\tan t}{\sqrt{t}}\). Both numerator \(\tan t\) and denominator \(\sqrt{t}\) go to \(0\), forming \(\frac{0}{0}\). Using L'Hôpital's Rule, differentiate the numerator and denominator: \(\frac{1}{\cos^2 t}\) and \(\frac{1}{2\sqrt{t}}\). As \(t\to 0^+\), this simplifies to \(\lim_{t\rightarrow 0^+} \frac{2\sqrt{t}}{\cos^2 t}\), which goes to \(0\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Infinity Limits

Infinity limits occur when one or both of the variables in a function get larger and larger without bound. This can be visualized as trying to reach beyond an unlimited distance. For many functions, this would cause both the numerator and denominator to increase indefinitely. Certain types of infinity limits require specific approaches to evaluate them effectively. To handle infinity limits, you often need specific strategies:

• Check if the numerator or denominator grows faster or at the same rate.
• If possible, simplify the expression to make it easier to evaluate its behavior at infinity.

In calculus, recognizing how forms approach infinity is crucial, because it helps decide on methods like L'Hôpital's Rule to make computations simpler.

Indeterminate Forms

When confronting limits, sometimes the expression reaches a form that does not definitively indicate any specific value. These are known as indeterminate forms. The most common indeterminate forms are \(\frac{0}{0}\) and \(\frac{\infty}{\infty}\), which require special strategies or rules for proper evaluation.Here are key takeaways about indeterminate forms:

• They require further analysis or transformation before a limit can be properly resolved.
• L'Hôpital's Rule is designed to help resolve these forms by differentiating the numerator and the denominator.

Do remember that just because an expression is indeterminate at first glance doesn’t mean the limit doesn't exist. Instead, it requires additional work to simplify or transform it for evaluation.

Limit Evaluation

Limit evaluation is key to understanding the behaviour of functions as they approach a specific point or infinity. Evaluating limits can reveal a function's trend or terminal value, a fundamental concept in calculus. Considerations while evaluating limits:

• Understand if the function produces any indeterminate forms.
• Use algebraic manipulations or rules like L'Hôpital's Rule.

Limits form the foundation of more advanced calculus topics such as continuity and differentiability. Handling them skillfully is essential for mastering calculus concepts. As illustrated in the examples, limits can be calculated directly or via rule-based transformations if forms appear indeterminate.

Differentiation

Differentiation, the process of finding a derivative, measures how a function changes at any point. This is especially useful in evaluating limits that involve expressions leading to indeterminate forms.In the context of limits:

• Differentiation transforms each part of \(\frac{\infty}{\infty}\) or \(\frac{0}{0}\) forms to facilitate a straightforward limit evaluation.
• L'Hôpital’s Rule specifically applies derivatives to deal with indeterminate forms, allowing ease in computation.

Accurate differentiation techniques improve the understanding of a function’s behavior as it approaches certain points. Calculating derivatives correctly is essential when applying methods such as L'Hôpital's Rule to solve limits problems efficiently.