Limits

Question

Evaluate the limit: $L = \lim_{x 	o 0} \frac{\ln(1 + x + x^2) + \ln(1 - x + x^2)}{\sec x - \cos x}$

RankGuru · Accepted Answer

Step 1: Simplify the numerator.
Using the property $\ln a + \ln b = \ln(ab)$, the numerator becomes:
$\ln((1+x^2)+x) + \ln((1+x^2)-x) = \ln((1+x^2)^2 - x^2) = \ln(1 + 2x^2 + x^4 - x^2) = \ln(1 + x^2 + x^4)$.

Step 2: Use the standard limit $\lim_{u 	o 0} \frac{\ln(1+u)}{u} = 1$.
Here $u = x^2 + x^4$. As $x 	o 0$, the numerator behaves like $x^2 + x^4 \approx x^2$.

Step 3: Simplify the denominator.
$\sec x - \cos x = \frac{1}{\cos x} - \cos x = \frac{1 - \cos^2 x}{\cos x} = \frac{\sin^2 x}{\cos x}$.
Using $\lim_{x 	o 0} \frac{\sin x}{x} = 1$, the denominator behaves like $\frac{x^2}{1} = x^2$.

Step 4: Calculate the final limit.
Ratio $= \frac{x^2}{x^2} = 1$.

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