Question

(i) Using L ' Hospital's rule find (a) $\underset{x\to 0}{lim}\frac{\tan x-x}{x^3}$ (b) $\underset{x\to 0}{lim}\frac{e^x-x-1}{x^2}$

• A. (i) (a) $\frac{1}{3}$ (b) $\frac{1}{2}$
• B. (i) (a) $\frac{1}{2}$ (b) $\frac{1}{3}$
• C. (i) (a) $\frac{2}{3}$ (b) $\frac{1}{4}$
• D. (i) (a) $\frac{1}{4}$ (b) $\frac{2}{3}$

Answer 1

Answer: (A)

(i) (a) $\frac{1}{3}$ (b) $\frac{1}{2}$

(a) $\underset{x\to 0}{lim}\frac{\tan x-x}{x^3}$

$=\underset{x\to 0}{lim}\frac{{\sec }^{2}x-1}{3x^2}$

$=\frac{1}{3}\underset{x\to 0}{lim}{\left(\frac{\tan x}{x}\right)}^2=\frac{1}{3}$

(b) $\underset{x\to 0}{lim}\frac{e^x-x-1}{x^2}$

$=\underset{x\to 0}{lim}\frac{e^x-1}{2x}$

$=\frac{1}{2}\underset{x\to 0}{lim}\frac{e^x-1}{x}=\frac{1}{2}$