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SAMPLE CALCULUS PROBLEMS

Problem 1. Find

Solution.

The problem presents the typical case of a quotient of two quantities that go to infinity. In fact, and both approach to infinity as . Which means that the quotient

is an undetermined form. Somehow we have to cancel the "bad" part, if possible. Notice that since both and are big, there's a chance for cancelation, so the quotient could be finite.

The trick in this case is to divide by the highest power on the expression (from both numerator and denominator). In this case we will divide both numerator and denominator by . We get

as because , and as .

This trick is applied in general for this type of expressions.

Problem 2. Find and if

Solution. Let's differentiate both sides of the equality. Let's recall that we are assuming that . That means that the equation gives us a way to solve as a function of , but we still don't know how to express that function. That assumption requires further justification (meaning, it's not always granted), but we'll take it for granted. So, since we get using the chain rule:

where . The 0 on the right hand side comes from differentiating the constant . We multiply the equation by and we get

which means that

Let's get . From the last equation we obtain easily that

and now we differentiate this again with respect to to get:

But and . We know put all the elements together:

and now we simplify the right hand side of the last equation

But by definition: , so we replace this into the previous equation: