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3 years ago

Lim x tanx/ 1-cosx x―>0 please help... ...?

1 answer:

8 0

Note that 1 - cos(x) = 2sin²(x/2)

⇒ lim x→0 xtan(x)/ [1 - cos(x)] =

= lim x→0 xtan(x) / 2sin²(x/2) =

= lim x→0 1/2 xtan(x) / [sin²(x/2) / (x/2)² * (x/2)²]

Also, we know that lim x→0 sin(x)/x = lim x→0 tan(x)/x = 1

So the limit is :

= 1/2 lim x→0 xtan(x) / (x²/4) =

= 1/2 lim x→0 4/x² * xtan(x) =

= 2 lim x→0 tan(x)/x =

<span>= 2.

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Step-by-step explanation:

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1 year ago

A rectangular Corn Hole area at the Recreation Center has a width of 5 feet and a length of 10 feet. If

lbvjy [14]

Answer:

2ft

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3 years ago

Find the tangent line equation of the curve at the given point. Y=arcsin(7x) at the point where x=sqrt2/14

Mumz [18]

Answer:

$Y-\frac{\pi}{2} =\frac{\pi}{2} (x+\sqrt{\frac{1}{7}})$

Step-by-step explanation:

The equation of the curve is

$Y = sin^{-1}(7x)$

To find the equation of tangent we need to differentiate this equation w.r.t x

So, differentiating we get

$Y'=\frac{7}{\sqrt{1-49x^2} }$

This would give the slope of the tangent line at any given point of which x coordinate is known. In the present case it is $x = \sqrt{\frac{1}{7} }$

Then slope would accordingly be

$Y'=\frac{7}{\sqrt{1-49/49} }$

= ∞

For, $x = \sqrt{\frac{1}{7} }$ , $Y = sin^{-1}(7/7)= \pi/2$

Equation of tangent line, in the point slope form, would be $Y-\frac{\pi}{2} =\frac{\pi}{2} (x+\sqrt{\frac{1}{7}})$

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3 years ago

A box contains 9 pounds of sand the entire box of sand will be poured in equal amounts into 6 small bags how many bags of sand w

spayn [35]

Answer: 1 1/2

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2 years ago