What is the limit of (sin(x)-tan(x))/x^3 when x-->0?

Answer 1

lim [ (sin(x) - tan(x) ) / x^3 )
x -> 0

To solve this, you have to use L'Hospital's rule. Note that plugging in x = 0, we will get [sin0 - tan0] / [0^3] = [ (0 - 0)/0^3 ] = 0/0, so we can safely use L'Hospital's rule.

Taking the derivative of the top and bottom, we have

lim [ (cos(x) - sec^2(x)) / (3x^2) ]
x -> 0

Note that sec^2(x) = 1/cos^2(x), so

lim [ (cos(x) - (1/cos^2(x)) ) / 3x^2 ]
x -> 0

Multiply top and bottom by cos^2(x) will eliminate all fractions within the fraction.

lim [ (cos^3(x) - 1) / [ (3x^2) cos^2(x) ]
x -> 0

Plugging in x = 0, we will get [ (1^3 - 1) / 0 ], or [0/0], so we use L'Hospitals' rule again.

lim [ ( 3cos(x)(-sin(x)) ) / [3 { (2x cos^2(x) + x^2 (2cos(x))(-sin(x)) }], x -> 0

lim [ (-3cos(x)sin(x) ) / [ 6x cos^2(x) - 6x^2 cos(x)sin(x) ]
x -> 0

At this point, we can cancel a single cos(x) from all the terms, giving us

lim [ (-3sin(x)) / [6x cos(x) - 6x^2 sin(x)]
x -> 0

At this point we have the form [0/(0 - 0)] which is still [0/0], so we use L'Hospitals' rule again. This time, let's pull out the constant -3/6.

(-3/6) lim [ sin(x) / [xcos(x) - x^2sin(x)] ]
x -> 0

Apply L'Hospital's rule,

(-1/2) lim [ cos(x) / [ cos(x) + x(-sinx) - [2xsin(x) + x^2cos(x)] ]
x -> 0

This is of the form [1 / (1 + 0 - 0 + 0)] = 1, so we have

(-1/2) (1)

And our answer is -1/2

Answer 2

1/2

Answer 3

Put tan as sin/cos.It comes to
sin(x) * [(cos(x)-1]/(cos(x) *x^3 =sin(x)/x *[cos(x)-1]x^2 *1/cos(x)

The first factor has limit 1.The second has limit -1/2 and the third has limit 1.

So the limit is -1/2

Answer 4

you may't take the logarithm of a destructive huge variety except you want to apply complicated numbers. in case you need to use complicated numbers then: logx( -36 ) = 2 Cancel the logarithm -36 = x² Take the sq. root 6i and -6i = x So one answer is 6i. in case you may like the respond to the project in polarized type, then the respond may well be: e^( ½i( pi - i * log( 36 ) ) ). i'm hoping this helps you. Have an magnificent day.