TD

They're asking for the possible values of "5x+2", not of x. So when you say 0 is a possibility, this doesn't mean "x = 0" but "5x+2 = 0". C is the correct answer.

The problem isn't the 4x, but the square! Since you can easily integrate sin(x), it's no problem to integrate sin(ax) with a an arbitrary constant either! To get rid of the square, use another trig identity: [math]\cos \left( {2x} \right) = \cos ^2 x - \sin ^2 x = 1 - 2\sin ^2 x \Leftrightarrow \sin ^2 x = \frac{{1 - \cos \left( {2x} \right)}}{2}[/math] Now you lost the square, went to a cosine and doubled the angle - but that shouldn't be a problem

I can't check your answer since the question isn't 100% clear to me. Is B located 6 miles under the road, vertically? So the distance between B and the road is 6. And the 10 miles, is that the (straight) distance between A and B or between A and the point on the road which is above B?

Ok Well in general for equations like this, we have that [math]\cos \alpha = \cos \beta \Leftrightarrow \alpha = \beta + 2k\pi \vee \alpha = - \beta + 2k\pi ,\forall k \in \mathbb{Z}[/math] I don't consider this as "intuitive" since this holds in general and is a perfect algebraic solution - but perhaps the topic starter can clarify what he meant.

What do you mean with non-intuive way? Two cosines are the same when either the arguments are the same or opposites, with 2k*pi of course.

You're on the right track. With your drawing, split the problem in 2 parts using an intermediate point P. Express the distance as a function of the distance to P by using the pyth. theorem on the triangle, compute the time necessary to do both parts with the given speeds and then add those two parts.

There's no need to apologize, but do you see what your mistake was? That's quite fundamental, in general (a+b)^n isn't a^n + b^n.

The answer is indeed 1/27 but be careful, b = 3, not -3 in this case (you substituted correctly though).

I'll go into a bit more detail. We would like to get rid off that cube root to simplify the numerator. For square roots, we can use the fact that [math]a^2 - b^2 = \left( {a - b} \right)\left( {a + b} \right)[/math] but for cube roots, that won't help. As I said, we'll be using the identity [math]\left( {a - b} \right)\left( {a^2 + ab + b^2 } \right) = a^3 - b^3[/math]. We consider the current numerator as the factor (a-b) with of course [math]a = \left( {x + 27} \right)^{1/3}[/math] and [math]b = 3[/math]. Now we multiply numerator and denominator with the same factor, namely the second one of our identity, so (a²+ab+b²) with our a and b. After doing that, instead of cancelling these equal factors (then we wouldn't have done a thing...) we can simplify the numerator since the expression there is now equal to a³-b³ according to our identity. But with our a and b, that becomes [math]a^3 - b^3 \Rightarrow \left( {\left( {x + 27} \right)^{1/3} } \right)^3 - 3^3 = x + 27 - 27 = x[/math] and the cube root is gone, just as we wished. The only thing that's left is an x, but that can be cancelled out with the x in the denominator leaving only our added expression in the denominator. It's now possible to simply fill in x = 0 in our limit to find the value.

Eeks! [math]\left( {a + b} \right)^{1/3} \ne a^{1/3} + b^{1/3} [/math] !!!

Use the fact that [math]\left( {a - b} \right)\left( {a^2 + ab + b^2 } \right) = a^3 - b^3 [/math] and consider the nominator as the factor (a-b) in that expression. Then find out what a and b are and multiply nominator & denominator by the second factor in that expression. The nominator now simplifies to a³-b³ and you'll be able to cancel out an x which will get rid off the indeterminate form. Then it's just filling in I basically said it all, now it's up to you to do the numbers!

That's true but I don't see the relevance of your complex example. Also in the reals, a limit isn't defined if it depends on the way you approach it (meaning: it has to be the same, coming from the left or from the right). In that way, the limit for x going to 0 of 1/x isn't defined because depending on coming from the left or right, you get - or + infinity.

Correct, with that nuance that it could be negative infinity as well of course -depending on the sign of the numerator and the way of approaching 0.

That's not correct. 0*(a finite number) is 0 but 0*inf is an indeterminate form, just as 0/0, inf/inf, inf-inf, ...