Solution to Exercise 8.1-3© 1998 by Karl Hahn |
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The problem was to find the limit of
arcsin(x) - π/2
lim
x → 1 √1 - x2
Step 1: Determine if both numerator and denominator go to zero.
We know that
arcsin(1) = π/2, so
clearly the numerator goes to zero as x approaches 1.
In the denominator, it's clear that x2 goes to
1 as x approaches 1. So the difference inside
the square root function must go zero. Square root is a continuous
function for nonnegative x, and the square root of zero is zero.
So as x approaches 1, the denominator must go to zero as well.
So this limit qualifies for L'Hopital's rule.
Step 2: Take the derivatives of the numerator and denominator. We discussed in section 7.2 how the derivative of arcsin(x) was
1
√1 - x2
Since the
π/2 is a constant, its derivative
is zero. So the derivative of the numerator is also
1
√1 - x2
To take the derivative of the denominator,
______
√1 - x2
you will need to use the
chain rule. When you do
you will get
1 -2x -xMake sure you can take this derivative on your own.=2 √1 - x2 √1 - x2
Step 3: Apply L'Hopital's Rule. That means
1
arcsin(x) - π/2 √1 - x2
lim = lim
x → 1 √1 - x2 x → 1 -x
√1 - x2
The
______
√1 - x2)
terms in the numerator and
denominator cancel, and you are left with
arcsin(x) - π/2 -1
lim = lim
x → 1 √1 - x2 x → 1 x
When you take the limit of -1/x as x approaches 1 you
get a limit of -1, which is the answer. You can experiment on your
calculator with taking the original function of x's that are very
close to 1 to confirm this limit.
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