Question

$\bold{Class}$: 11th

$\bold{Subject}$: Maths

$\bold{Lesson}$: Limits and Derivatives

Solve...

Answer 1

❤ Saiyyan ❤

Hey niku ❤,

Just see the attachment... ♥

♥ Saiyyan ♥

Answer 2

According to the Question

$\bf\huge\boxed{\boxed{\bf\huge\:Hello\:Mate}}}$

≈ $\bf\huge\frac{sin\:6x}{x} - \frac{sin\:9x}{x}$

≈ $\bf\huge\frac{sin\:6x}{x}\times \frac{6}{6} - \frac{sin\:9x}{x}\times \frac{9}{9}$

≈ $\bf\huge\frac{sin\:6x}{x}\times 6 - \frac{sin\:9x}{9x}\times9$

= 6 - 9

= -3

Second :-

⇒ $\bf\huge\frac{8\:sin\:theta}{theta}\times Theta - \frac{Theta\:cos\:Theta}{theta}\times Theta$ ÷ $\bf\huge\frac{3\:tan \:Theta}{Theta}\times Theta + Theta^2$

⇒ $\bf\huge\frac{8\:Theta-Theta^2}{3\:Theta- Theta^2}$

⇒ $\bf\huge\frac{Theta(8-Theta)}{Theta(3-Theta)}$

⇒ $\bf\huge\frac{8-Theta}{3-Theta}$

≈ $\bf\huge\frac{8}{3}$

$\bf\huge\boxed{\boxed{\bf\huge\:Solved}}}$