Solution to Exercise 8.1-4© 1998 by Karl Hahn |
The problem was to find the limit of
lim x sin(1/x) x → ∞Step 1: Express this as a quotient. So you have
lim x sin(1/x) = x → ∞ |
lim x → ∞ |
sin(1/x) |
Step 2: Confirm that both numerator and denominator go to zero. As x gets very large, 1/x gets very close to zero, and indeed goes to zero in the limit. So the denominator goes to zero. And since we are taking sin(1/x) and we have determined that 1/x goes to zero, and since sine is a continuous function, it must be the case that sin(1/x) goes to zero as well as x grows without limit (i.e. as x goes to infinity). So this limit qualifies for L'Hopital's Rule.
Step 3: Take the derivatives of the numerator and denominator. The denominator is easy. Its derivative is -1/x^{2}. The function, sin(1/x), is a composite, so we have to use the chain rule to find its derivative. When you do, you find that its derivative is
cos(1/x) (-1/x^{2})
Step 4: Apply L'Hopital's Rule. That means
lim x → ∞ |
sin(1/x) |
lim x → ∞ |
-1 cos(1/x) |
Step 5: Take the limit. The -1/x^{2} terms in the numerator and denominator cancel. You are left with cos(1/x). We know that as x grows without limit, 1/x goes to zero. Since cosine is a continuous function, the limit as x grows without limit of cos(1/x) must be cos(0) = 1, which is the answer.
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