lim_(y->0)(sin^2(3y)-3y^2)/(y^2+y^3) = ?

2 Answers
Dec 4, 2017

Supposing that you mean:

lim_(y rarr 0) (sin^2(3y) - 3y^2)/(y^2 + y^3)

The answer is 6.

Explanation:

I will run through two methods: the 'unofficial' method, and the (more) 'official' method.

Intuitive approach around the limit:

Using the compound angle and double angle formulas for sin and cos:

sin(3y) = sin (2y + y)
=sin(2y)cosy + cos(2y)siny)
=2sinycos^2y+(2cos^2y-1)siny
=4sinycos^2y - siny

(Rearranging to get it all in terms of sin(y))

=4siny(1-sin^2y) - siny
=3siny - 4sin^3y

Now we want to find the square of that (sin(3y)):

sin^2(3y) = (3siny - 4sin^3y)^2
=(siny(3-4sin^2y))^2
=sin^2y(3- (4 - 4cos^2y))^2
= sin^2y(4cos^2y - 1)^2

Now to find:

lim_(y rarr 0) (sin^2y(4cos^2y - 1)^2 - 3y^2)/(y^2 + y^3)

This is where we start to get a bit 'intuitive'...

As y rarr 0, siny rarr 0
rArr siny rarr y
Hence let y = siny.

Meanwhile cosy rarr 1
Hence let cosy = 1.

This simplifies the limit down to:

lim_(y rarr 0) (y^2(3)^2 - 3y^2)/(y^2 + y^3)
=lim_(y rarr 0)(6y^2)/(y^2 + y^3)
=6 lim_(y rarr 0) y^2/(y^2(1+y))
=6 lim_(y rarr 0) 1/(1+y)

Finally letting y = 0 we get:

6 lim_(y rarr 0) 1
= 6

Official approach to the limit:

This uses L'Hopital's Rule, i.e. if lim_(y rarr 0) f(x)/g(x) yields a mathematically impossible result, then the limit will be the same if you repeatedly differentiate f and g to the point where the result is no longer undefined.

lim_(y rarr 0) (sin^2(3y) - 3y^2)/(y^2 + y^3)

lim_(y rarr 0) (d/dx(sin^2(3y) - 3y^2))/(d/dx(y^2 + y^3))

By the chain rule, this is equal to:

lim_(y rarr 0) (6sin(3y)cos(3y) - 6y)/(2y + 3y^2)

Still gives out the zero denominator, so we do it all again!

By the product and chain rules:

d/dx(6sin(3y)cos(3y)
=6(3cos(3y)cos(3y) + (-sin(3y)sin(3y))
=6(3cos^3y - 3sin^3y)

Hence:

lim_(y rarr 0) (6sin(3y)cos(3y) - 6y)/(2y + 3y^2)
=lim_(y rarr 0) (18cos^2(3y) - 18sin^2(3y) - 6)/(2 +6y)

Rearranging the trigonometric parts of the numerator using the Pythagorean Identity, we get:

lim_(y rarr 0) (36cos^2(3y) - 18 - 6)/(2 +6y)

Finally letting y=0, we get:

(36-18-6)/2
=6

Goodness, that was long - I hope it helps!

Dec 4, 2017

6

Explanation:

(sin^2(3y)-3y^2)/(y^2+y^3) = (((sin^2(3y))/y^2)-3)/(1+y) = (9((sin^2(3y))/(3y)^2)-3)/(1+y) =( 9( (sin(3y))/(3y))^2-3)/(1+y)

and now knowing that lim_(y->0) (sin(3y))/(3y) = 1

lim_(y->0)(sin^2(3y)-3y^2)/(y^2+y^3) =( 9( lim_(y->0)(sin(3y))/(3y))^2-3)/(1+lim_(y->0)y) = (9-3)/1 = 6