Question

Plz prove fast as possible......

Answer 1

Answer:

Step-by-step explanation:

Lim(x→ 0) { tan2x - sin2x }/x³  

check which form of limit apply here,  

put x = 0  

0/0 hence, we can apply L- HOSPITAL rule ,  

Lim(x→0) {tan2x - sin2x}/x³  

differentiate separately numerator and denominator wrt x .

= Lim(x→0) { 2sec²2x - 2cos2x}/3x²  

= 2/3× Lim(x→0) { 1 - cos³2x }/cos²2x.x²  

= 2/3 × Lim(x→0) {(1 -cos2x)(1 + cos²2x +cos2x}/cos²2x.x²  

= 2/3 × Lim(x→0){ (2sin²x/x²)(1+cos²2x + cos2x)}/cos²2x  

we know,  

Lim(f(x)→0) sinf(x)/f(x) = 1  

= 4/3 × Lim(x→0) {(sinx/x)² × (1+cos²2x + cos2x)/cos²2x }  

= 4/3 × 1 × ( 1 + 1 + 1)/1  

= 4/3 × 1 × 3  

= 4 ( answer )

that's the answer

Answer 2

tan2x =2tanx/1-tan^2x
sin2x=2tanx /1+tan^2x
lim x-0 tan2x-sin2x /x^3
lim x-0 2tanx/1-tan^2x-2tanx/1+tan^2x/x^3
please calculted