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Alona [7]
3 years ago
6

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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Answer:

Step-by-step explanation:

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6 0
3 years ago
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Rainbow [258]

Answer:

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Step-by-step explanation:

3 0
3 years ago
If logb2=x and logb3=y, evaluate the following in terms of x and y:
Alja [10]

log_b{162} = x + 4y\\\\log_b324 = 2x+4y\\\\log_b\frac{8}{9} = 3x-2y\\\\\frac{log_b27}{log_b16} = 3y-4x

<em><u>Solution:</u></em>

Given that,

log_b2 = x\\\\log_b3 = y --------(i)

<em><u>Use the following log rules</u></em>

Rule 1: log_b(ac) = log_ba + log_bc

Rule 2: log_b\frac{a}{c} = log_ba - log_bc

Rule 3: log_ba^c = clog_ba

a) log_b{162}

Break 162 down to primes:

162 = 2^1 \times 3^4

log_b{162} =log_b 2^1. 3^4\\\\By\ rule\ 1\\\\ log_b{162} = log_b 2^1 +log_b 3^4\\\\By\ rule\ 3\\\\1log_b2 + 4log_b3\\\\1x+4y\\\\x+4y

Thus we get,

log_b162 = x + 4y

Next

b) log_b 324

Break 324 down to primes:

324 = 2^2 \times 3^4

log_b324 = log_b 2^2.3^4\\\\By\ rule\ 1\\\\log_b324 = log_b2^2 + log_b3^4\\\\By\ rule\ 3\\\\log_b324 = 2log_b2 + 4log_b3\\\\From\ (i)\\\\log_b324 = 2x + 4y

Next

c) log_b\frac{8}{9}

By rule 2

log_b\frac{8}{9} = log_b8 - log_b9\\\\log_b\frac{8}{9} = log_b 2^3 - log_b3^2\\\\By\ rule\ 3\\\\log_b\frac{8}{9} =  3 log_b2 - 2log_b3\\\\From\ (i)\\\\log_b\frac{8}{9} =  3x - 2y

Next

d) \frac{log_b27}{log_b16}

By rule 2

\frac{log_b27}{log_b16} = log_b27 - log_b16\\\\ \frac{log_b27}{log_b16} = log_b3^3 - log_b2^4\\\\By\ rule\ 2\\\\ \frac{log_b27}{log_b16} = 3log_b3 - 4log_b2 \\\\From\ (i)\\\\\frac{log_b27}{log_b16} =  3y - 4x

Thus the given are evaluated in terms of x and y

3 0
2 years ago
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Verizon [17]

Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:they are because when you simplify they are the same

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