Question

Lim x~0 Sin(x) cos (x) / akar phi + 2sin(x) - akar phi

Answer 1

Example: limx 0 x → 2
If we apply l’Hˆopital’s rule to this problem we get:
sin x cos x lim = lim (l’Hop) x 0 x → 2 x→0 2x
= lim − sin x (l’Hop) x→0 2
= 0.
If we instead apply the linear approximation method and plug in sin x ≈ x,
we get:
sin x x
x2 ≈ x2
1 ≈ . x
We then conclude that:
sin x lim = x 0+ x2 ∞ →
sin x
x
lim
0− x2 = −∞. →
There’s something fishy going on here. What’s wrong?
Student: L’Hˆopital’s rule wasn’t applied correctly the second time. cos x 0 That’s correct; limx 0 2x is of the form 1
0 , not 0 → or some other indetermi-
nate form.
This is where you have to be careful when using l’Hˆopital’s rule. You have
to verify that you have an indeterminate form like 0
0 or ∞
∞ before applying the
rule. The moral of the story is: Look before you l’Hˆop.
Also, don’t use l’Hospital’s rule as a crutch. If we want to evaluate:
x5 − 2x4 + 1 limx→∞ x4 + 2
we can apply l’Hˆopital’s rule four times, or we could divide the numerator and
denominator by x5 to conclude:
x5 − 2x4 + 1 1 − 2/x + 1/x5
lim = lim x→∞ x4 + 2 x→∞ 1/x + 2/x5
1 = 0 = ∞.
After enough practice with rates of growth, we can calculate this limit almost
instantly:
lim x5 − 2x4 + 1 lim x5
= ∞. x4 + 2 ∼ x x→∞ x→∞ 4
1

Answer 2

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