# lim(tanx-1)/[(sinx)^2 - (cosx)^2]   x->pi/4

hala718 | High School Teacher | (Level 1) Educator Emeritus

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Let f(x0 = (tanx -1)/(sin^2 x - cos^2 x)

lim (tanx - 1) /[ sin^2 x - cos^2 x)     x --> pi/4

First let us substitute with x = pi/4

==> lim f(x) = ( 1-1)/(1/2 - 1/2) = 0/0 ( the method failed)

But we know that: tanx = sinx/cosx

==< lim f(x) = lim (sinx/cosx -1)/(sin^2 x - cos^2 x)

= lim (sinx - cosx)/cosx(sin^2 x - cos^2 x)

= lim (sinx - cosx)/ cosx(sin x-cosx )(sinx + cosx)

= lim 1/cosx(sinx+cosx)

==> lim f(x) x---> pi/4 = 1/cospi/4(cospi/4 + sinpi/4)

= 1/(sqrt2/2)*(sqrt2)

= 2/2= 1

Then lim f(x) when x--> pi/4 =  1

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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First, we'll verify if it is a case of indeterminacy, so we'll substitute x by pi/4, into the given expression:

f(pi/4) = (tan pi/4-1)/[(sin pi/4)^2 - (cos pi/4)^2]  , where tan pi/4= 1, cos pi/4  =sin pi/4  =sqrt2/2

f(pi/4) = (1-1)/(1/2 - 1/2)

f(pi/4) = 0/0,  indeterminacy case

We'll solve the limit, by substituting the difference

tan x - 1 = (sin x / cos x) - 1 = (sin x - cos x)/ cos x

lim(tanx-1)/[(sinx)^2 - (cosx)^2] = lim (sin x - cos x)/ cos x* [(sinx)^2 - (cosx)^2]

We'll write the difference of square from denominator as:

[(sinx)^2 - (cosx)^2] = (sinx-cosx)(sinx+cosx)

lim (sin x-cos x)/ cos x*[(sinx)^2-(cosx)^2]=lim (sin x-cos x)/cosx*(sinx-cosx)(sinx+cosx)

We'll reduce like terms,(sin x-cos x):

lim 1/cosx*(sinx+cosx) =1/cos pi/4*(sin pi/4 + cos pi/4)

lim 1/cosx*(sinx+cosx) = 2/(sqrt2)(sqrt2)

lim 1/cosx*(sinx+cosx) = 2/2

lim 1/cosx*(sinx+cosx) = 1

neela | High School Teacher | (Level 3) Valedictorian

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lim(tanx-1)/[(sinx)^2 - (cosx)^2} as x -->pi/4.

Solution:

At x= pi/4, (tanx-1)/(sin^2-cos^2) becomes (1-1)/(1/2 -1/2)  is 0/0 form of inditermination, as  when x= pi/4,  sinx = cos x = 1/sqrt2.

So we use L'Hopital's rule of diffrentiating numerator and denominator and  then take the linit asx-->pi/4.

Lt(tanx-1)/(sin^2x-cos^2x) = lt (tanx-1)'/(sin^2x-cos2x)'....(1)

RHS of (1)  =   Lt(asx-->pi/4)  {sec^2x}/{2sinxcosx- 2cosx(-sinx)} = 1/cosx* 4sinxcosx = 1/{4 (1/sqrt2)^2(1/sqrt2)(1/sqrt)} = 4/4 =1, as sinp/4 = cospi/4 =  1/sqrt2

So Lt(tanx-1)/ (sin^2x- cos^2x) = 1.

Second method:

(tanx -1) /(sin^2 - cos^2x) = (sinx/cosx -1)(sin^2-cos^2x)

(tanx-1)/(sin^2x-cos^2x) = (sinx -cosx)/{cosx (sinx+cosx)(sinx+cosx)}

(tanx-1)/(sin^2-cos^x) = 1/{cosx(sinx+cosx)}

Now we take the limit asx--> pi/4.

Lt(tanx-1)/(sin^2x - cos^2 x) = Lt 1/ {cosx(sinx+cosx)

lt(tanx-1)/(sin^2-cos^2x) = 1/{cospi/4(sipi/2+cospi/4)} = 1{1/sqrt)(1/sqrt2+1/sqrt2)} = 1/ {(1/sqrt2)(2/sqrt2)} = 1/(2/2) = 1

Therefore lt(tanx-1)/(sin^2x-cos^2x) = 1 as x--> pi/4.