Grade 12IIT JEE Entrance Exam

Find the value of the given Lim x tending to 0 cos(tanx) - cosx/x^4

Himanshu yadav

7 Years agoGrade 12

7 Years ago

Let L be the given limit.

$\therefore$ L = lim _x0[cos(tanx) – cosx] / x⁴ ...Using cosA – cosB = 2 sin{(A + B)/2} sin{(A – B)/2}

= lim _x0 [2 sin{(tanx + x)/2} sin{(x – tanx)/2}] / x⁴

Multiply and divide by {(tanx + x)/2} {(x – tanx)/2}

L = 2 . lim _x0 sin{(tanx + x)/2} . {(tanx + x)/2} / (tanx + x)/2 .

sin{(x – tanx)/2} . {(x – tanx)/2} / {(x – tanx)/2} / x⁴

As x $\rightarrow$ 0, (tanx + x)/2 $\rightarrow$ 0 and (x – tanx)/2 $\rightarrow$ 0.

L = 2 . lim _x0 sin{(tanx + x)/2} / (tanx + x)/2 .lim _x0 sin{(x – tanx)/2} / {(x – tanx)/2} .

lim _x0{(tanx + x)/2} . {(x – tanx)/2} / x⁴

We know that lim _x0 tanx / x = 1

$\therefore$ lim _x0 sin{(tanx + x)/2} / (tanx + x)/2 = 1

and lim _x0 sin{(x – tanx)/2} / {(x – tanx)/2} = 1

So, L = 2 . 1 . 1 . lim _x0{(tanx + x) / 2} . {(x – tanx) / 2} / x⁴

= 2 lim _x0(tanx + x) . (x – tanx) / 4x⁴

= ½ lim _x0 (x² – tan²x) / x⁴

Here use expansion of tangent series.

tanx = x + (1/3) x³ + (2/15) x⁵ + ….

tan²x = x² + (2/3) x⁴ + f(x), where degree of x in f(x) is greater than 4.

$\therefore$ L = ½ lim _x0 [ x² – {x² + (2/3) x⁴ + f(x)} ] / x⁴

L = ½ lim _x0 [ x² – x² – (2/3)x⁴ – f(x)} ] / x⁴

L = ½ lim _x0 [– (2/3) x⁴ – f(x)} ] / x⁴

L = ½ lim _x0 [–(2/3) x⁴ / x⁴ ] – ½ lim _x0 [ f(x) / x⁴ ] = (1/2) (– 2/3) = – 1/3

$\because$ f(x) contains powers of x greater than 4, lim _x0 [ f(x)} ] / x⁴ = 0,

$\therefore$ lim _x0[cos(tanx) – cosx] / x⁴ = – 1/3.