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prohojiy [21]
3 years ago
15

Given limit f(x) = 4 as x approaches 0. What is limit 1/4[f(x)]^4 as x approaches 0?

Mathematics
1 answer:
stepladder [879]3 years ago
3 0

Answer:

\displaystyle 64

General Formulas and Concepts:

<u>Calculus</u>

Limits

Limit Rule [Variable Direct Substitution]:                                                             \displaystyle \lim_{x \to c} x = c

Limit Rule [Variable Direct Substitution Exponential]:                                         \displaystyle \lim_{x \to c} x^n = c^n

Limit Property [Multiplied Constant]:                                                                     \displaystyle \lim_{x \to c} bf(x) = b \lim_{x \to c} f(x)

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

\displaystyle  \lim_{x \to 0} f(x) = 4

<u>Step 2: Solve</u>

  1. Rewrite [Limit Property - Multiplied Constant]:                                           \displaystyle \lim_{x \to 0} \frac{1}{4}[f(x)]^4 = \frac{1}{4} \lim_{x \to 0} [f(x)]^4
  2. Evaluate limit [Limit Rule - Variable Direct Substitution Exponential]:       \displaystyle \lim_{x \to 0} \frac{1}{4}[f(x)]^4 = \frac{1}{4}(4^4)
  3. Simplify:                                                                                                         \displaystyle \lim_{x \to 0} \frac{1}{4}[f(x)]^4 = 64

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Limits

Book: College Calculus 10e

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Step-by-step explanation:

\huge\begin{array} {|c|} \hline\left. \begin{cases} {   \rm\frac{ 1  }{ 4  }  x+ \frac{ 4  }{ 5  }  y =  3  } \\ {   \rm\frac{ 5  }{ 16  }  x- \frac{ 1  }{ 5  }  y=5  }  \end{cases} \right. \\  \hline \end{array}

\\ \Large \overline{ \rm -   \: Steps \:  using  \: substitution \: - }

\rm \frac{1}{4}x+\frac{4}{5}y=3

\rm \frac{1}{4}x=-\frac{4}{5}y+3

\rm x=4\left(-\frac{4}{5}y+3\right)

\rm x=-\frac{16}{5}y+12

┈

\rm \frac{5}{16}\left(-\frac{16}{5}y+12\right)-\frac{1}{5}y=5

\rm -y+\frac{15}{4}-\frac{1}{5}y=5

\rm -\frac{6}{5}y+\frac{15}{4}=5

\rm -\frac{6}{5}y=\frac{5}{4}

\bold{ y=-\frac{25}{24}}

┈

\rm x=-\frac{16}{5}\left(-\frac{25}{24}\right)+12

\rm x=\frac{10}{3}+12

\bold{ x=\frac{46}{3}}

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