Question

Arrange the following limits in ascending order
a)$\underset{x\to 0}{lim}\frac{\sin x^0}{x}$
b)$\underset{x\to 0}{lim}\frac{3\sin x^0-\sin 3x^0}{x^3}$
c)$\underset{x\to 0}{lim}\frac{tan{x}^0}{2x}$
d)$\underset{x\to 0}{lim}\frac{(\sec x^0+1)(\sec x^0-1)}{x^2}$

• A. b,d,a,c
• B. b,c,a,d
• C. b,c,d,a
• D. b,a,d,c

Answer 1

The correct option is A b,d,a,c

Using $x=\frac{\pi }{180}\left(x^0\right)$
a)$\underset{x\to 0}{lim}\frac{\sin x^0}{x}=\underset{x\to 0}{lim}\frac{\sin x^0}{\frac{\pi }{180}{x}^0}=\frac{180}{\pi }$
b)$\underset{x\to 0}{lim}\frac{3\sin x^0-\sin 3x^0}{x^3}=\underset{x\to 0}{lim}\frac{4{\sin }^3{x}^0}{(\pi /180x^0{)}^3}=4{\left(\frac{180}{\pi }\right)}^3$
c)$\underset{x\to 0}{lim}\frac{\tan x^0}{2x}=\underset{x\to 0}{lim}\frac{\tan x^0}{2\left(\pi /180\right)x^0}=\frac{90}{\pi }$
d)$\underset{x\to 0}{lim}\frac{(\sec x^0+1)(\sec x^0-1)}{x^2}=\underset{x\to 0}{lim}\frac{({\sec }^2{x}^0-1)}{(\pi /180x^0{)}^2}=\underset{x\to 0}{lim}\frac{{\tan }^2{x}^0}{(\pi /180x^0{)}^2}={\left(\frac{180}{\pi }\right)}^2$
Hence ascending order of limits is, b,d,a,c