How do you determine the limit algebraically?

Mathematics

1 answer:

eduard3 years ago

5 0

Answer:

0.125

Step-by-step explanation:

This is a standard calculus question. If you search out L'Hopital's Rule you will find that there are only Calculus examples present. If you use calculus, the answer comes to 1/8. The only way I can think of doing this without the use of Calculus is to let x = something very small -- say x = 0.0001 That approaches 0, but it is not quite = to 0.

So let's try it.

16 + 0.00001 = 16.00001

sqrt(16.0001) = 4.00000125

Now subtract 4

4.00000125 - 4

0.00000125

Now divide by 0.00001

0.12499998

which is close to 0.125.

If someone else can give you an answer that is "purer" than this one, take it.

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Elden [556K]

Answer: a) $\bold{\dfrac{3}{16}}$ b) $\bold{\dfrac{1}{36}}$

Step-by-step explanation:

a) In order to get an even number, you have the 3 different scenarios:

1) Even, Even, Even, Even $\dfrac{3\times 3\times 3\times 3}{6^4} \quad = \dfrac{3^4}{6^4}\quad =\dfrac{1}{16}$

2) Even, Even, Odd, Odd $\dfrac{3\times 3\times 3\times 3}{6^4} \quad = \dfrac{3^4}{6^4}\quad =\dfrac{1}{16}$

3) Odd, Odd, Odd, Odd $\dfrac{3\times 3\times 3\times 3}{6^4} \quad = \dfrac{3^4}{6^4}\quad =\dfrac{1}{16}$

Order doesn't matter

Add them up to get your answer: $\dfrac{1}{16}+\dfrac{1}{16}+\dfrac{1}{16}\quad =\large\boxed{\dfrac{3}{16}}$

b) If one die is a 2 and another is a 3 and the other two dice can be any number, then you have 1 possibility for a 2, 1 possibility for a 3, and 6 possibilities for each of the other two dice.

$\dfrac{1\times 1\times 6\times 6}{6^4}\quad =\dfrac{1}{6^2}\quad =\large\boxed{\dfrac{1}{36}}$