# What is the correct option of the following question?

Feb 9, 2018

The answer is $B$

#### Explanation:

For this problem I assume that $\log \left(a\right) = \ln \left(a\right)$

Let's rewrite the denominator in terms of sine and cosine exclusively.

$\sec x - \cos x = \frac{1}{\cos} x - \cos x = \frac{1 - {\cos}^{2} x}{\cos} x = {\sin}^{2} \frac{x}{\cos} x$

We can also use the laws of logarithms to simplify the numerator to

$\ln \left[\left(1 + x + {x}^{2}\right) \left(1 - x + {x}^{2}\right)\right]$

Now let's rewrite the limit.

$L = {\lim}_{x \to 0} \frac{\ln \left[\left(1 + x + {x}^{2}\right) \left(1 - x + {x}^{2}\right)\right]}{{\sin}^{2} \frac{x}{\cos} x}$

$L = {\lim}_{x \to 0} \frac{\cos x \ln \left[\left(1 + x + {x}^{2}\right) \left(1 - x + {x}^{2}\right)\right]}{\sin} ^ 2 x$

If we try to evaluate the limit now, we will get $\frac{0}{0}$, so we can apply L'Hospital's rule.

Finding the derivative of the numerator may be a little bit long. We must first differeintaite $\ln \left[\left(1 + x + {x}^{2}\right) \left(1 - x + {x}^{2}\right)\right]$ using the chain rule.

Start by expanding within the brackets:

$1 - x + {x}^{2} + x - {x}^{2} + {x}^{3} + {x}^{2} - {x}^{3} + {x}^{4}$

Now combine like terms

${x}^{4} + {x}^{2} + 1$

Now differentiate using the chain and power rules.

$\frac{d}{\mathrm{dx}} \left(\ln \left({x}^{4} + {x}^{2} + 1\right)\right) = \frac{4 {x}^{3} + 2 x}{{x}^{4} + {x}^{2} + 1}$

Now use the product rule to differentiate the entire numerator.

$L = {\lim}_{x \to 0} \frac{- \sin x \ln \left({x}^{4} + {x}^{2} + 1\right) + \cos x \left(\frac{4 {x}^{3} + 2 x}{{x}^{4} + {x}^{2} + 1}\right)}{\sin \left(2 x\right)}$

You will find that this gives $\frac{0}{0}$ once more. We will have to differentiate again. Repeat this process until you get something you can evaluate (not being $\frac{0}{0}$).

It's tedious, but in the end you should get an answer of $1$. The graph confirms:

Hopefully this helps!