What is the correct option of the following question?

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1 Answer
Feb 9, 2018

Answer:

The answer is #B#

Explanation:

For this problem I assume that #log(a) = ln(a)#

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

#secx - cosx= 1/cosx - cosx = (1 -cos^2x)/cosx = sin^2x/cosx#

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

#ln[(1 + x + x^2)(1 - x + x^2)]#

Now let's rewrite the limit.

#L = lim_(x-> 0) (ln[(1 +x +x^2)(1 -x + x^2)])/(sin^2x/cosx)#

#L = lim_(x->0) (cosxln[(1 +x + x^2)(1 - x+ x^2)])/sin^2x#

If we try to evaluate the limit now, we will get #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[(1 + x + x^2)(1 - x + x^2)]# 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.

#d/dx(ln(x^4 + x^2 + 1)) = (4x^3 + 2x)/(x^4 + x^2 + 1)#

Now use the product rule to differentiate the entire numerator.

#L = lim_(x->0) (-sinxln(x^4 +x^2 + 1) + cosx((4x^3 + 2x)/(x^4 + x^2 + 1)))/(sin(2x))#

You will find that this gives #0/0# once more. We will have to differentiate again. Repeat this process until you get something you can evaluate (not being #0/0#).

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

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Hopefully this helps!