Question

$\underset{x\to 0}{lim}\frac{\log (1+x+x^2)+\log (1-x+x^2)}{\sec x-\cos x}=$

• A. $1$
• B. $-1$
• C. $0$
• D. $\infty$

Answer 1

Answer: (A)

$1$

$\underset{x\to 0}{lim}\frac{\log (1+x+x^2)+\log (1-x+x^2)}{\sec x-\cos x}$

$=\underset{x\to 0}{lim}\frac{\log \left[\right(1+x+x^2\left)\right(1-x+x^2\left)\right]}{\frac{1}{\cos x}-\cos x}$

$=\underset{x\to 0}{lim}\frac{\cos x\log \left[\right(1+x^2{)}^2-x^2]}{1-{\cos }^{2}x}$

$=\underset{x\to 0}{lim}\frac{\cos x\log [1+x^2+x^4]}{{\sin }^{2}x}$

$=\underset{x\to 0}{lim}\cos x\cdot \frac{\log [1+x^2(1+x^2\left)\right]}{x^2(1+x^2)}\cdot (1+x^2)\cdot {\left(\frac{x}{\sin x}\right)}^2$

$=1\cdot 1\cdot 1\cdot 1=1$