I need a step-by-step solution for this limit

Question

I need the limit of f(x)=(6sin(x )^2-sqrt(2)sin(x )-2)/(tan(x )-1) as x->π/4

Please don’t use derivates, I haven’t studied them yet. The result should be 5/2, but I don’t know how to get it.

Thanks a lot!

Answer 1

Let's first plug in the pi/4 and see what we get: 0/0 is indeterminate, so we need to find a way to factor the top or rationalize the top or do SOMETHING to get something other than 0/0.

So my first step was to LET u = sin(x) Now I can represent the numerator as 6u^2 -√2 u - 2

I used the quadratic formula and found that u = √2/2 and u = -√2/3

Turning these into factors, I get 2u - √2= 0 and 3 u + √2 = 0 or (2u - √2)(3u + √2)

And as the last step for the numerator I replace the "u"s with sin(x)

So our numerator becomes

[ 2sin(x) - √2 ] [ 3sin(x) + √2 ]

Now we turn to the denominator.

If we multiply (tanx - 1) by (tanx + 1) = tan^2(x) - 1 (and we must also multiply the numerator by (tanx + 1) )

tan^2(x) can be re-written using the half-angle formulas as

1 - cos(2x)

---------------

1 + cos(2x)

so tan^2(x) - 1 = (replacing 1 with [ 1 + cos(2x) ]/[ 1 + cos(2x) ] )

1 - cos(2x) [ 1 + cos(2x) ]

--------------- - -----------------------

1 + cos(2x) [ 1 + cos(2x) ]

And this becomes

1 - cos(2x) - 1 - cos(2x) -2 cos(2x)

------------------------------- = ---------------------------------

1 + cos(2x) 1 + cos(2x)

Re-capping: the denominator tan^2x - 1 can be replaced with

-2 cos(2x)

--------------

1 + cos(2x)

NOW - let's put the fraction back together

[ 2 sin(x) - √2 ] [ 3sin(x) +√2 ] [ tan(x) + 1 ]

--------------------------------------------------------- and this can be re-written as

-2 cos(2x)

----------------------

1 + cos(2x)

[ 1 + cos(2x) ] [ 2sin(x) - √2 ] [ 3sin(x) + √2 ] [ tan(x) + 1 ]

----------------------------------------------------------------------------

-2 cos(2x)

Using double-angle formula, cos(2x) = 1 - 2 sin^2(x) which factors into [ 1 - √2 sin(x) ] [ 1 + √2 sin(x) ]

Factor a -1 out of the first factor and this becomes (-1) [ √2 sin(x) - 1 ] [ 1 + √2 sin(x) ]

Then factor out a √2, and we get -√2[ sin(x) - 1/√2 ] [ 1 + √2 sin(x)]

And the denominator of -2 cos(2x) becomes -2 [ -√2 ] [ sin(x) - 1/√2 ] [1 + √2 sin(x) ]

And last simplification of the denominator gives us: 2 √2 [ sin(x) - 1/√2 ] [ 1 + √2 sin(x) ]

Look at one of the middle terms in the numerator, the [ 2 sin(x) - √2 ] term. We factor a 2 out and get

2 [ sin(x) - √2 /2 ]

But √2/2 = 1/√2 so that middle term in the numerator can be written 2 [ sin(x) - 1/√2 ]

NOW put it all together:

[ 1 + cos(2x) ] (2) [ sin(x) - 1/√2 ] [ 3sin(x) + √2 ] [ tanx + 1 ]

------------------------------------------------------------------------------

2√2[ sin(x) - 1/√2 ] [ 1 + √2 sin(x)]

Now we have a [ sin(x) - 1/√2 ] and a 2 on both the top and the bottom and they will both cancel out!

[ 1 + cos(2x) ] [ 3sin(x) + √2 ] [ tanx + 1 ]

-------------------------------------------------------

√2 [ 1 + √2 sin(x) ]

NOW find the limit as x→pi/4:

[ 1 + cos(pi/2) ] [ 3 sin(pi/4) + √2 ] [ tan(pi/4) + 1 ]

----------------------------------------------------------------

√2 [ 1 + √2 sin(pi/4) ]

[1+0] [ 3/√2 + √2 ] [1 + 1] [1] [ 3/√2 + 2/√2 ] [2] [ 5/√2][2] 5 √2

---------------------------------- = ---------------------------------- = ---------- = -------

√2 [ 1 + √2(1/√2) ] √2 [ 1 + 1 ] √2 [2] 2 √2

= 5/2

And that was a ton of work!!!!!

Answer 2

What's the value of "x"?

Is "π"=3.14159263.............. or is it 180° (radians°)?