Question

The value of $\underset{x\to 0}{lim}\frac{ln(1+k{x}^2)}{1-cosx}$ is

• A. $0$
• B. $1$
• C. $k$
• D. $2k$

Answer 1

Answer: (D)

$2k$

$\underset{x\to 0}{lim}\frac{ln(1+k{x}^2)}{1-cosx}=k\underset{x\to 0}{lim}\frac{\frac{ln(1+k{x}^2)}{k{x}^2}}{\frac{1-cosx}{x^2}}$
$=k\underset{x\to 0}{lim}\frac{\frac{ln(1+k{x}^2)}{k{x}^2}}{\frac{2{\sin }^2\frac{x}{2}}{x^2}}=2k\underset{x\to 0}{lim}\frac{\frac{ln(1+k{x}^2)}{k{x}^2}}{{\left(\frac{\sin \frac{x}{2}}{\frac{x}{2}}\right)}^2}=2k$