Editorial - Matlan - Fungsi Trigonometri

Some Insights

• focus on the terms that tend to 0

• always rationalize if the square roots are the culprits

• only substitute x for something else when it is needed

• if all else fails then find the limit in parts

• use t and not y because it's clearer when you write it

• refer to some useful trigonometric identities when needed

Problem 1

(lim_x→π)((1+cos(x))/(sin^2)(x))

This is just simple identity substitution.

(sin^2)(x)+(cos^2)(x)=1

(sin^2)(x)=1-(cos^2)(x)=(1+cos(x))*(1-cos(x))

(lim_x→π)((1+cos(x))/((1+cos(x))*(1-cos(x))))=(lim_x→π)(1/(1-cos(x)))=1/(1-(-1))=1/2

Problem 2

(lim_x→π/4)((1-tan(x))/cos(2*x))

Identity substitution.

cos(2*x)=(1-(tan^2)(x))/(1+(tan^2)(x))

Use the difference of squares.

a2-b2=(a-b)*(a+b)

(lim_x→π/4)((1-tan(x))⋅(1+(tan^2)(x))/(1-(tan^2)(x)))=(lim_x→π/4)((1-tan(x))⋅(1+(tan^2)(x))/((1+tan(x))*(1-tan(x))))

(lim_x→π/4)((1+(tan^2)(x))/(1+tan(x)))=(1+12)/(1+1)=1

Problem 3

(lim_x→0)((tan^2)(3*x)/(1-cos(2*x)))

Identity substitution.

cos(2*x)=1-2*(sin^2)(x)

(lim_x→0)((tan^2)(3*x)/(2*(sin^2)(x)))=1/2⋅3⋅3=9/2

Problem 4

(lim_x→0)((2*x*tan(3*x))/(cos(2*x)-cos(3*x)))

Identity substitution.

cos(a)-cos(b)=-2*sin((a+b)/2)*sin((a-b)/2)

(lim_x→0)((2*x*tan(3*x))/(-2*sin((5*x)/2)*sin(-x/2)))=(lim_x→0)(2/(-2)⋅x/sin((5*x)/2)⋅tan(3*x)/sin(-x/2))

-1⋅2/5⋅3⋅(-2)=12/5

Problem 5

(lim_x→0)((tan(3*x)-sin(3*x))/(2*x3))=(lim_x→0)((sin(3*x)/cos(3*x)-(sin(3*x)*cos(3*x))/cos(3*x))/(2*x3))

(lim_x→0)((sin(3*x)-sin(3*x)*cos(3*x))/(2*x3*cos(3*x)))=(lim_x→0)((sin(3*x)*(1-cos(3*x)))/(2*x3*cos(3*x)))

Identity substitution.

cos(0)=1

cos(a)-cos(b)=-2*sin((a+b)/2)*sin((a-b)/2)

1-cos(3*x)=-2*sin((3*x)/2)*sin(-(3*x)/2)=2*(sin^2)((3*x)/2)

(lim_x→0)((sin(3*x)*(2*(sin^2)((3*x)/2)))/(2*x3*cos(3*x)))

Separate the cos(3*x).

2/2⋅3⋅3/2⋅3/2⋅1/cos(0)=27/4

Problem 6

(lim_x→0)(2/(x2)-cot(3*x)/(x2*csc(6*x)))=(lim_x→0)(2/(x2)-(cos(3*x)/sin(3*x)⋅sin(6*x))/(x2))

This is just simple identity substitution.

sin(2*x)=2*sin(x)*cos(x)

(lim_x→0)((2-cos(3*x)/sin(3*x)⋅2*sin(3*x)*cos(3*x))/(x2))=(lim_x→0)((2-2*(cos^2)(3*x))/(x2))

(lim_x→0)((2*(sin^2)(3*x))/(x2))=2⋅3⋅3=18

Problem 7

(lim_x→2*π)((sin(x)-sin(2*π))/(2*π-x))

t substitution makes this trivial.

x-2*π=0=t

x=t+2*π

(lim_t→0)((sin(t+2*π)-sin(2*π))/(2*π-t-2*π))=(lim_t→0)((sin(t)-0)/(-t))=-1

Problem 8

(lim_x→3*π/4)((sin(x)+cos(x))/cos(2*x))

Not every problem needs a t substitution, especially this early.

Identity substitution.

cos(2*x)=(cos^2)(x)-(sin^2)(x)=(cos(x)+sin(x))*(cos(x)-sin(x))

(lim_3*π/4)((sin(x)+cos(x))/((cos(x)+sin(x))*(cos(x)-sin(x))))

(lim_3*π/4)(1/(cos(x)-sin(x)))=1/(-√(,2))=-√(,2)/2

Problem 9

(lim_x→5*π/4)((5*π-4*x)/(cot(x)-tan(x)))

For this one, 5*π-4*x is basically begging for a t substitution.

x-(5*π)/4=0=t

x=t+(5*π)/4

(lim_t→0)((5*π-4*(t+(5*π)/4))/(cot(t+π/4)-tan(t+π/4)))

Let's work on the deonominator first.

cot(t+π/4)-tan(t+π/4)=(1-tan(t))/(tan(t)+1)-(tan(t)+1)/(1-tan(t))

((1-tan(t))2-(1+tan(t))2)/(1-(tan^2)(t))=-(4*tan(t))/(1-(tan^2)(t))=-2⋅(2*tan(t))/(1-(tan^2)(t))

-2*tan(2*t)

(lim_t→0)((-4*t)/(-2*tan(2*t)))=(lim_t→0)(2⋅t/tan(2*t))=2⋅1/2=1

Problem 10

(lim_x→11*π/6)((1+2*sin(x))/(11*π-6*x))

t substitution from the start.

x-(11*π)/6=0=t

x=t+(11*π)/6

(lim_t→0)((1+2*sin(t+(11*π)/6))/(-6*t))

Substitute using identity.

2*sin(t+(11*π)/6)=2*(sin(t)*cos((11*π)/6)+cos(t)*sin((11*π)/6))

√(,3)*sin(t)-cos(t)

(lim_t→0)((1-cos(t))/(-6*t)+(√(,3)*sin(t))/(-6*t))

Another identity.

1-cos(t)=2*(sin^2)(t/2)

(lim_t→0)((2*(sin^2)(t/2))/(-6*t)+√(,3)/(-6))

I can't lie this solution is kind of weird.

(2*(sin^2)(t/2))/(-6*t)=(sin(t/2)/t/2)2⋅t/12

The limit of that is 12⋅0=0

The 0 factors out which means that this is a determinant form.

(lim_t→0)(0+√(,3)/(-6))=-√(,3)/6

Problem 11

(lim_x→1/3)((18*x2-39*x+11)/sin(3*x-1))

Trust the process and use t substitution.

x-1/3=0=t

x=t+1/3

(lim_t→0)((18*(t+1/3)2-39*(t+1/3)+11)/sin(3*t))=(lim_t→0)((18*t2-27*t)/sin(3*t))

(lim_t→0)((9*t(2*t-3))/sin(3*t))=9/3⋅(0-3)=-9

Problem 12

(lim_x→π/2)((2*x-π)/(3*(2*x-π)-tan(2*x-π)))

Yet another t substitution.

x-π/2=0=t

x=t+π/2

(lim_t→0)(t/(3*t-tan(t)))=(lim_t→0)((t*cos(t))/(3*t*cos(t)-sin(t)))

(lim_t→0)(cos(t)/(3*cos(t)-sin(t)/t))

sin(t)/t approaches 1 as t approaches 1

(lim_t→0)(cos(t)/(3*cos(t)-1))=1/(3-1)=1/2

Problem 13

(lim_x→0)((tan(5*x)*(sin^2)(4*x)-8*x3)/(3*x2*tan(4*x)*cos(3*x)))

Looks complicated, but this can be easily solved by solving limits within limits.

Divide numerator and denominator by x3

(lim_x→0)(((tan(5*x)*(sin^2)(4*x))/(x3)-8)/(3*tan(4*x)*cos(3*x))/x)

Solve the limits within limits.

(5⋅4⋅4-8)/(3⋅4⋅1)=72/12=6

Remember that cosines don't really matter in limits at 0 because their value is 1 there.

Problem 14

(lim_x→0)((cos(2*x)*(sin^2)(3*x)+(cos^2)(3*x)-1)/(2*x4))

Looks complicated, but we can just use identities.

(cos^2)(3*x)-1=-(sin^2)(3*x)

(lim_x→0)((cos(2*x)*(sin^2)(3*x)-(sin^2)(3*x))/(2*x4))

(lim_x→0)(((sin^2)(3*x)*(cos(2*x)-1))/(2*x4))

Yet another identity used.

cos(2*x)=1-2*(sin^2)(x)

cos(2*x)-1=-2*(sin^2)(x)

(lim_x→0)(((sin^2)(3*x)*(-2*(sin^2)(x)))/(2*x4))=-2/2⋅3⋅3⋅1⋅1=-9

Problem 15

(lim_x→4)(((x2-4*x)*tan(2*x-8))/((x2-2*x-8)*(√(,2*x2-6*x+1)-3)))

Rationalize.

(lim_x→4)((x*(x-4)*tan(2*x-8)*(√(,2*x2-6*x+1)+3))/((x-4)*(x+2)⋅2*(x+1)*(x-4)))

Split the limit.

(lim_x→4)(tan((2*(x-4)))/(x-4)⋅x/(x+2)⋅(√(,2*x2-6*x+1)+3)/(2*(x+1))⋅(x-4)/(x-4))

2⋅4/6⋅(3+3)/(2⋅5)=2⋅2/3⋅3/5=12/15=4/5

Problem 16

(lim_x→∞)(2*x2*sin(3/x)*tan(4/x))

t substitution as always.

x=∞=1/t

1/x=1/∞=0=t

(lim_t→0)(2⋅1/(t2)⋅sin(3*t)⋅tan(4*t))

2⋅3⋅4=24

Wow!

Problem 17

(lim_x→3)((x2-3*x)/sin(3-√(,3*x)))

Weird t substitution.

t=3-√(,3*x)

3*x=(t-3)2=t2-6*t+9

x-3=-2*t+(t2)/3

(lim_x→3)(x(x-3)/sin(3-√(,3*x)))

Calculation hell

(lim_t→0)(((3-2*t+(t2)/3)*(-2*t+(t2)/3))/sin(t))

(lim_t→0)(((-6+5*t-(4*t2)/3+(t3)/9)*t)/sin(t))

1⋅(-6)=-6

Not so complicated after all

Problem 18

(lim_x→3*π/2)((cos^2)(x)/((2*x-3*π)2*sin(x)))

Yet another t substitution done early.

x-(3*π)/2=t=0

x=t+(3*π)/2

(lim_t→0)((sin^2)(t)/(-4*t2*cos(t)))=(lim_t→0)((tan(t)*sin(t))/(-4*t2))=-1/4

Easy

Problem 19

(lim_x→π/2)(((π-2*x)2)/(√(,5-sin(x))-2))

t substitution as always.

x-π/2=t

x=t+π/2

(lim_t→0)((4*t2)/(√(,5-cos(t))-2))

Rationalize.

(lim_t→0)((4*t2*(√(,5-cos(t))+2))/(1-cos(t)))

Another usage of this identity.

1-cos(t)=2*(sin^2)(t/2)

(lim_t→0)((4*t2*(√(,5-cos(t))+2))/(2*(sin^2)(t/2)))

This is trivial to solve now.

(lim_t→0)((4*t2)/(2*(sin^2)(t/2))⋅(√(,5-cos(t))+2))

2⋅2⋅2⋅(2+2)=8⋅4=32

Problem 20

yeah the t substitution is unnecessary here. this method is unnecessarily long.

(lim_x→2)(((x2-y2)*sin(x-2))/(x3+(y-2)*x2-2*x*y))

Yet another t substitution.

x-2=0=t

x=t+2

y-2=h

y=h+2

(lim_x→2)((((t+2)2-(h+2)2)*sin(t))/((t+2)3+h⋅(t+2)2-2*(t+2)*(h+2)))

wtf is this

Numerator

(x-y)*(x+y)*sin(t)

(t+2-h-2)*(t+2+h+2))sin(t)

(t-h)*(t+h+4)*sin(t)

Denominator

t3+6*t2+12*t+8+t2*h+4*t*h+4*h-2*(t*h+2*t+2*h+4)

t3+6*t2+12*t+8+t2*h+4*t*h+4*h-2*t*h-4*t-4*h-8

t3+6*t2+t2*h+8*t+2*t*h

(t2+(6+h)*t+2*h+8)*t

Finally, the end

(lim_t→0)(((t-h)*(t+h+4))/(t2+(6+h)*t+2*h+8)⋅sin(t)/t)

Substitute t=0

(-h(h+4))/(2*h+8)

Substitute h=y-2

-((y-2)*(y+2))/(2*(y-2)+8)=-(y2-4)/(2*y+4)=-((y-2)*(y+2))/(2*(y+2))

-(y-2)/2=(2-y)/2

yay

Problem 21

(lim_x→5*π/4)((3*(1-sin(2*x)))/((4*x-5*π)*sin(12*x-15*π)))

t substitute again

x-(5*π)/4=t

x=t+(5*π)/4

(lim_t→0)((3*(1-sin(2*t+(5*π)/2)))/(4*t*sin(12*t)))

(lim_t→0)((3*(1-cos(2*t)))/(4*t*sin(12*t)))

Yet another use of this identity.

1-cos(t)=2*(sin^2)(t/2)

(lim_t→0)((6*(sin^2)(t))/(4*t*sin(12*t)))=6/4⋅1/12=1/8

Problem 22

(lim_x→3*π/4)((1+2*sin(x)*cos(x))/(sin(x)+cos(x)))

This one doesn't need t substitution.

We have to recognize this identity, though.

Let p=sin(x)+cos(x)

p2=(sin^2)(x)+(cos^2)(x)+2*sin(x)*cos(x)=1+2*sin(x)*cos(x)

Therefore transforming this problem into:

(p2)/p=p

(lim_x→3*π/4)(sin(x)+cos(x))=0

Problem 23

(lim_x→0)((2*x*(cos(8*x)-1))/(√(,sin(2*x)+4)-√(,tan(2*x+4))))

Feels like a rationalization is needed.

(lim_x→0)((2*x*(cos(8*x)-1))/(sin(2*x)-tan(2*x))⋅(√(,sin(2*x)+4)+√(,tan(2*x)+4)))

Identity transformation.

(sin^2)(4*x)=(1-cos(8*x))/2

-2*(sin^2)(4*x)=cos(8*x)-1

(lim_x→0)((-4*x*(sin^2)(4*x))/(sin(2*x)-tan(2*x))⋅(…))

Trust the process?

(lim_x→0)((-4*x*(sin^2)(4*x)*cos(2*x))/(sin(2*x)*cos(2*x)-sin(2*x))⋅(…))

Because I'm lazy I'll put this here

sin(2*x)*cos(2*x)-sin(2*x)

sin(2*x)*(cos(2*x)-1)

sin(2*x)*(-2*(sin^2)(x))

-2*sin(2*x)*(sin^2)(x)

Also:

(sin^2)(4*x)=4*(sin^2)(2*x)*(cos^2)(2*x)

(lim_x→0)((-16*x*(sin^2)(2*x)*(cos^3)(2*x))/(-2*sin(2*x)*(sin^2)(x))⋅(…))

(lim_x→0)(8*x⋅sin(2*x)/(sin^2)(x)⋅(cos^3)(2*x)⋅(…))

Yet another usage of this identity.

sin(2*x)=2*sin(x)*cos(x)

(lim_x→0)((16*x)/sin(x)⋅cos(x)⋅(cos^3)(2*x)⋅(2+2))=16⋅1⋅13⋅4=64

Problem 24

(lim_x→0)((6*x+3*sin(2*x)-7*x*cos(2*x))/(2*x2+2*tan(3*x)-sin(2*x)))

divide by x

(lim_x→0)((6+(3*sin(2*x))/x-7*cos(2*x))/(2*x+(2*tan(3*x))/x-sin(2*x)/x))u

(6+6-7)/(0+6-2)=5/4

Problem 25

(lim_x→1)((sin(x-1)-tan(x-1))/((x2-2*x+1)*sin(x-1)))

t substitution.

x-1=t

t+1=x

x2-2*x+1=(x-1)2

(lim_t→0)((sin(t)-tan(t))/(t2*sin(t)))

(lim_t→0)((sin(t)*cos(t)-sin(t))/(t2*sin(t)*cos(t)))=(lim_t→0)((cos(t)-1)/(t2*cos(t)))

(lim_t→0)((-2*(sin^2)(t/2))/(t2)⋅1/cos(t))

-2/2⋅1/2⋅1/1=-1/2

Problem 26

(lim_x→0)((3*x3+2*(tan^2)(4*x)-(2*sin(3*x))2)/(7*x2-10*x+4*(sin^2)(3*x)+10*x*cos(3*x)))

Divide by x2

(lim_x→0)((3*x+(2*(tan^2)(4*x))/(x2)-(4*(sin^2)(3*x))/(x2))/(7+(4*(sin^2)(3*x))/(x2)+(10*(cos(3*x)-1))/x))

I don't want to repeat myself too much.

cos(3*x)-1=-2*(sin^2)(3/2*x)

(-20*(sin^2)(3/2*x))/x

-20⋅(sin(3/2)/(3/2*x))2⋅9/4*x

x→0

then this expression also →0

(0+2⋅4⋅4-4⋅3⋅3)/(7+4⋅3⋅3+0)=(32-36)/(7+36)=-4/43

Annoying

Problem 27

(lim_5*x→4)(sin(5/2*π*x)/(2-√(,5*x)))

5*x=4

x=4/5

(lim_x→4/5)(sin(5/2*π*x)⋅(2+√(,5*x))/(4-5*x))

Now we can substitute for t

x-4/5=0=t

t+4/5=x

(lim_t→)(sin(5/2*t*π+2*π))*(2+√(,5*t+4))/(-5*t)

(lim_x→4/5)(sin(5/2*t*π)/(-5*t)⋅(2+√(,5*t+4)))=(5*π)/2⋅1/(-5)⋅4=-2*π

Problem 28

(lim_x→π/4)(((π-4*x)*tan(4*x)*cos(10*x))/(cot(6*x)-cot(6*x)*sin(10*x)))

This should be straightforward.

x-π/4=t

x=t+π/4

(lim_t→0)((-4*t*tan(4*t+π)*cos(10*t+(5*π)/2))/(cot(6*t+(3*π)/2)-cot(6*t+(3*π)/2)*sin(10*t+(5*π)/2)))

(lim_t→0)((4*t*tan(4*t)*sin(10*t))/(-tan(6*t)+tan(6*t)*sin(10*t)))

(lim_t→0)((4*t*tan(4*t)*sin(10*t))/(tan(6*t)*(cos(10*t)-1)))

Just guessing here.

(sin^2)(5*t)=(1-cos(10*t))/2

-2*(sin^2)(5*t)=cos(10*t)-1

(lim_t→0)((4*t*tan(4*t)*sin(10*t))/(-2*tan(6*t)⋅(sin^2)(5*t)))

We have 3 on top and 3 on the bottom.

(lim_t→0)(-2⋅t/tan(6*t)⋅tan(4*t)/sin(5*t)⋅sin(10*t)/sin(5*t))

-2⋅1/6⋅4/5⋅10/5=-16/30=-8/15

Not so hard after all.

Problem 29

(lim_x→0)(1/(x*tan(x))-(cos^2)(x)/(x*sin(x)))

Weirdly easy?

(lim_x→0)(1/(x*tan(x))-cos(x)/(x*tan(x)))

(lim_x→0)((1-cos(x))/(x*tan(x)))

IDENTITY USED.

1-cos(x)=2*(sin^2)(x/2)

(lim_x→0)((2*(sin^2)(x/2))/(x*tan(x)))=2⋅1/2⋅1/2=1/2

Problem 30

(lim_x→π/2)((cos^2)(x)/((cos^2)(x)+(cot^2)(x)))

Simple. Don't need to substitute for t here.

(lim_x→π/2)(((cos^2)(x)*(sin^2)(x))/((sin^2)(x)*(cos^2)(x)+(cos^2)(x)))

Divide all by (cos^2)(x)

(lim_x→π/2)((sin^2)(x)/((sin^2)(x)+1))=1/(1+1)=1/2

:D