Given the functions  and ,

(a)    Evaluate h(g(2)).

(b)    If , f(1) can only exist if g(1) = h(1). Find a so that f(1) exists.

Difficulty: 

Solution

(a)    To evaluate this expression, substitute 2 into g (in place of x) and then substitute the result into h:

(b)    This is a bit trickier. We know that g(1) = h(1) from the problem. Finding g(1) is easy:

Now, we know that h(1) = g(1), so h(1) = 2. In other words, if we plug in 2 for x in the expression called h, the result must be 2. This equation is very easy to set up, and finding a is as natural as chewing on tree bark (for those who find such things natural, that is).

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Evaluate the following delicious little limit: .

Solution

One of the fundamental properties of limits is that the limit of a sum is equal to the sum of the limits. In other words, you can evaluate this limit by calculating each individual term's limit as x approaches 0 and then adding up the results. You can evaluate the limit of the last two terms by direct substitution, but to do the first two, you have to remember these special limit cases: . You'll have to multiply the first term by 2/2 to get the denominator to be 2x (to match the numerator's x term and thus be able to apply the special limit property). In the second term, you can factor out a 1/3 to get the x terms to match:

You can further justify this answer by examining the graph of that funky mess. It clearly is heading toward a height of 3 when x = 0.

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