Posted by Soroban on August 28, 2002 at 17:31:50:
In Reply to: very quick help with limits posted by nick on August 27, 2002 at 20:58:10:
Ryan's advice is correct: L'Hopital is the way to go.
But from the nature of the problems, though, I suspect that L'Hopital
is NOT the method they expect. In fact, the problems look like they
come from a section *before* L'Hopital is taught.
Can we find the limits anyway? Yes!
Two of them require knowledge of this theorem:
lim (sin z)/z = 1 (as z -> 0)
You probably saw it first when differentiating the sine function -
and thought (or hoped) you'd never see it again. *LOL*
You'll be surprised at how it can be used. (I remember that *I* was.)
: lim (sin x)/(2x²-x) [x -> 0]
Factor the denominator and write the expression as: (sin x)/x(2x - 1)
Then write it as two fractions: [(sin x)/x](1/(2x-1)]
Take the limit [x -> 0]:
lim (sin x)/x * lim[1/(2x-1)]
The first limit is 1, the second is -1:
Answer: (1)(-1) = -1
: lim (3sin 4x)/(sin 3x) [x -> 0]
This one takes some acrobatics.
we have: (3 sin 4x)/(sin 3x)
Divide numerator and denominator by 3x:
[(sin 4x)/x]/[(sin 3x)/3x]
In the numerator, multiply by 4/4:
Now, we take the limit [x -> 0]:
4 * lim[(sin 4x)/4x] / lim[(sin 3x)/3x]
Both limits equal 1. Answer: 4*1/1 = 4
: lim (1/(2+x)-1/2)/x [x -> 0]
This one requires only algebra.
Simplify the numerator (get a common denominator and combine):
1/(2+x) - 1/2 = -x/[2(2+x)]
Write it over x and cancel:
-x/[2(2+x)]/x = -1/[2(2+x)
The limit [x -> 0] is -1/4
We have to be careful when using that theorem:
lim (sin z)/z = 1 [ z -> 0]
Note that the "z" in the sine function is
the SAME as the "z" in the denominator.
You're thinking "Well, duh!", but bear with me.
If we have: (sin 2x)/x, the "z" is NOT the same.
We need a "2x" in the denominator.
So, we multiply by 2/2: 2(sin 2x)/2x
And NOW we can take the limit: (sin 2x)/2x -> 1,
so the limit is 2(1) = 2.
Got it? It's a tricky thing to digest.
And it should explain the "acrobatics" in the second problem.
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