121616949-math.97 - respectively zero 1 and ∞ We can in...

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4.8 Limits revisited 83 EXAMPLE 4.14 Compute lim x 0 sec x - 1 sin x . Both the numerator and denominator approach zero, so applying L’Hˆopital’s Rule: lim x 0 sec x - 1 sin x = lim x 0 sec x tan x cos x = 1 · 0 1 = 0 . EXAMPLE 4.15 Compute lim x 0 + x ln x . This doesn’t appear to be suitable for L’Hˆopital’s Rule, but it also is not “obvious”. As x approaches zero, ln x goes to -∞ , so the product looks like (something very small) · (something very large and negative). But this could be anything: it depends on how small and how large . For example, consider ( x 2 )(1 /x ), ( x )(1 /x ), and ( x )(1 /x 2 ). As x approaches zero, each of these is (something very small) · (something very large), yet the limits are
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Unformatted text preview: respectively zero, 1, and ∞ . We can in fact turn this into a L’Hˆopital’s Rule problem: x ln x = ln x 1 /x = ln x x-1 . Now as x approaches zero, both the numerator and denominator approach infinity (one-∞ and one + ∞ , but only the size is important). Using L’Hˆopital’s Rule: lim x → + ln x x-1 = lim x → + 1 /x-x-2 = lim x → + 1 x (-x 2 ) = lim x → +-x = 0 . One way to interpret this is that since lim x → + x ln x = 0, the x approaches zero much faster than the ln x approaches-∞ ....
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