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MatroZZZ [7]
3 years ago
13

Solve the following equation: lim x->0 sin3x/5x^3-4x.

Mathematics
1 answer:
goblinko [34]3 years ago
3 0
Usual limit of sin is sinX/X--->1, when X--->0

sin3x/5x^3-4x=0/0?, sin3x/3x--->1 when x --->0, so sin3x/5x^3-4x=                    [3x. sin3x / 3x] /(5x^3-4x)=(sin3x / 3x) . (3x/5x^3-4x)
   =(sin3x / 3x) . (3/5x^2- 4)
finally lim  sin3x/5x^3-4x=lim (sin3x / 3x) .(3/5x^2- 4)=1x(3/-4)= - 3/4
                 x----->0            x---->0
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Answer:

2

Step-by-step explanation:

d.

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Step-by-step explanation:

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Step-by-step explanation:

Two key things you need to know to solve this is that the lines will be parallel if their slopes are the same, and perpendicular if one slope is the negative reciprocal of the other (i.e. s_{1} = -s_{2}^{-1})

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s = \Delta y / \Delta x\\s = (12 - 2) / (5 - 0)\\s = 10 / 5\\s = 2

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3 years ago
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