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Chapter 2: Problem 50

Evaluate each limit and justify your answer. $$\lim _{t \rightarrow 2} \frac{t^{2}+5}{1+\sqrt{t^{2}+5}}$$

Short Answer

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Question: Evaluate the limit and justify your answer: $$\lim _{t \rightarrow 2} \frac{t^{2}+5}{1+\sqrt{t^{2}+5}}$$ Answer: The limit exists and is equal to \(\frac{9}{4}\). This is because when we substitute \(t = 2\) into the expression, we get a real number and the denominator is not zero.

Step by step solution

01

Evaluate the denominator

Let's first substitute the value of \(t\) to find the denominator. We have to make sure it is not zero. If it is zero, then we cannot find a limit. Denominator: \(1 + \sqrt{t^2 + 5}\) For \(t = 2\), \(1 + \sqrt{2^2 + 5} = 1 + \sqrt{9} = 1 + 3 = 4 \neq 0\)
02

Direct substitution

Since the denominator is nonzero at \(t = 2\), we can substitute this value directly into the given expression: $$\frac{t^{2}+5}{1+\sqrt{t^{2}+5}} = \frac{2^{2}+5}{1+\sqrt{2^{2}+5}} = \frac{4+5}{1+\sqrt{9}} = \frac{9}{1+3} = \frac{9}{4}$$
03

Justify the answer

Finally, we found the limit as a finite number, \(\frac{9}{4}\). This limit exists because when we plug in \(t = 2\), the value of the expression is a real number and the denominator is not zero. So, we have: $$\lim _{t \rightarrow 2} \frac{t^{2}+5}{1+\sqrt{t^{2}+5}} = \frac{9}{4}$$

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Key Concepts

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

Direct Substitution
Direct substitution is one of the simplest methods for finding limits in calculus. It involves directly substituting the value that the variable approaches into the function. In the given problem, where we evaluate \( \lim _{t \rightarrow 2} \frac{t^{2}+5}{1+\sqrt{t^{2}+5}} \), we begin by substituting \( t = 2 \) into the expression. This step is straightforward and efficient when the function is continuous at that point.

Why is direct substitution so handy? Here鈥檚 why:
  • It often avoids complicated algebraic manipulations.
  • If the function's denominator is non-zero and the function is well-defined, this method works perfectly.
  • It provides a direct path to the limit value.
Once confirmed that the function doesn鈥檛 have a zero denominator at the point of interest, direct substitution gives us the value \( \frac{9}{4} \), as the problem illustrates.
Denominator Evaluation
Before performing direct substitution, it's crucial to check if the denominator of the expression equals zero at the point we're considering. If the denominator is zero, the limit cannot be evaluated simply by substitution, and further techniques are needed. In the expression \( \frac{t^{2}+5}{1+\sqrt{t^{2}+5}} \), we first examine the denominator by replacing \( t \) with 2.

Let's break down why this step matters:
  • If a denominator becomes zero, dividing by zero leads to an undefined situation which requires special consideration, such as factoring or rationalizing, to resolve it.
  • Checking the denominator can prevent errors by ensuring the substitution is legitimate.
For \( t = 2 \), the denominator evaluates to 4, a non-zero value, making direct substitution a valid approach for finding the limit.
Limit Justification
Justifying the limit is the final assurance that we've correctly evaluated it. After performing direct substitution and ensuring the denominator isn't zero, our result should be a well-defined real number. In this case, after substituting \( t = 2 \), the expression simplifies neatly to \( \frac{9}{4} \).

Why bother with justification?
  • It confirms the meticulous steps were followed and the calculations are sound.
  • It ensures that all assumptions about the continuity and behavior of the function were correct.
  • It solidifies understanding and provides clear reasoning for the obtained limit.
Since the final value is a real number and our expression didn't encounter issues like division by zero, we can confidently say that the limit exists, adding this to our mathematical toolkit with confidence.

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Most popular questions from this chapter

Evaluate the following limits, where \(c\) and \(k\) are constants. \(\lim _{h \rightarrow 0} \frac{100}{(10 h-1)^{11}+2}\)

Determine whether the following statements are true and give an explanation or counterexample. a. The value of \(\lim _{x \rightarrow 3} \frac{x^{2}-9}{x-3}\) does not exist. b. The value of \(\lim _{x \rightarrow a} f(x)\) is always found by computing \(f(a)\) c. The value of \(\lim _{x \rightarrow a} f(x)\) does not exist if \(f(a)\) is undefined. d. \(\lim _{x \rightarrow 0} \sqrt{x}=0\) \(\lim _{x \rightarrow \pi / 2} \cot x=0\)

Steady states If a function \(f\) represents a system that varies in time, the existence of \(\lim f(t)\) means that the system reaches a steady state (or equilibrium). For the following systems, determine whether a steady state exists and give the steady-state value. The population of a culture of tumor cells is given by \(p(t)=\frac{3500 t}{t+1}\)

a. Use the identity \(\sin (a+h)=\sin a \cos h+\cos a \sin h\) with the fact that \(\lim _{x \rightarrow 0} \sin x=0\) to prove that \(\lim _{x \rightarrow a} \sin x=\sin a\) thereby establishing that \(\sin x\) is continuous for all \(x\). (Hint: Let \(h=x-a\) so that \(x=a+h\) and note that \(h \rightarrow 0\) as \(x \rightarrow a\).) b. Use the identity \(\cos (a+h)=\cos a \cos h-\sin a \sin h\) with the fact that \(\lim _{x \rightarrow 0} \cos x=1\) to prove that \(\lim _{x \rightarrow a} \cos x=\cos a\).

a. Create a table of values of \(\tan (3 / x)\) for \(x=12 / \pi, 12 /(3 \pi), 12 /(5 \pi) \ldots . .12 /(11 \pi) .\) Describe the general pattern in the values you observe. b. Use a graphing utility to graph \(y=\tan (3 / x) .\) Why do graphing utilities have difficulty plotting the graph near \(x=0 ?\) c. What do you conclude about \(\lim _{x \rightarrow 0} \tan (3 / x) ?\)
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