Page 76 - Calculus Demystified

CHAPTER 2
and Foundations of Calculus 63
lim h(x) =
x→c
then
lim f(x) = .
x→c
EXAMPLE 2.5
2
3
Calculate lim x→3 4x − 7x + 5x − 9.
SOLUTION
We may apply Theorem 2.1(a) repeatedly to see that
3
2
2
3
lim 4x − 7x + 5x − 9 = lim 4x − lim 7x + lim 5x − lim 9. (∗)
x→3 x→3 x→3 x→3 x→3
We next observe that lim x→3 x = 3. This assertion is self-evident, for when x
is near to 3 then x is near to 3. Applying Theorem 2.1(d) and Theorem 2.1(b)
repeatedly, we now see that

3
lim 4x = 4 · lim x · lim x · lim x = 4 · 3 · 3 · 3 = 108.
x→3 x→3 x→3 x→3
Also
2
lim 7x = 7 · lim x · lim x = 7 · 3 · 3 = 63,
x→3 x→3 x→3

lim 5x = 5 · lim x = 5 · 3 = 15.
x→3 x→3
Of course lim x→3 9 = 9.
Putting all this information into equation (∗) gives
3 2
lim 4x − 7x + 5x − 9 = 108 − 63 + 15 − 9 = 51.
x→3
EXAMPLE 2.6
Use the Pinching Theorem to analyze the limit

lim x sin x.
x→0
SOLUTION
We observe that

−|x|≡ g(x) ≤ f(x) = x sin x ≤ h(x) ≡|x|.
Thus we may apply the Pinching Theorem. Obviously

lim g(x) = lim h(x) = 0.
x→0 x→0
We conclude that lim x→0 f(x) = 0.