unit 1 differential calculus - IGNOU

(vi)
lim
x →a
[ lim f ( x)]
⎡ f ( x)
⎤ x →a
⎢ ⎥ =
provided lim g(
x)
≠ 0.
⎣ g(
x)
⎦ [ lim g ( x)]
x →a
x →a
Using the above properties, we try a few examples.
Example 1.11
Solution
Evaluate
2
lim ( x
x → 2
Solution
−
Example 1.12
2
lim ( x
x →2
4)
( x
2
3
−
4)
( x
2
+ x + 1)
+ x + 1)
= [ lim ( x
x + 2x
+ x
Evaluate lim
.
x →0
2
x + 2x
First we reduce
Hence,
x
3
lim
x
2
3
2
2
x → 2
2
− 4)
] [ lim ( x + x + 1)]
x → 2
= (4 – 4) (4 + 2 + 1) = 0.
+ 2x
+ x
by cancelling the common factor :
2
x + 2x
+ 2x
+ x x(
x + 1)
( x + 1)
= =
2
x + 2x
x(
x + 2)
x + 2
x
3
2
+ 2x
+ x ( x + 1)
= lim
x + 2x
x →0
2
x →0 2
x +
2
lim ( x + 1)
x →0
1
=
= .
lim ( x + 2)
2
x →0
We now turn to another limit, the importance of which will become clear in this
unit.
Theorem 3
sin x
(i) = 1.
(ii)
lim
x →0 x
lim
x →0
cos x = 1.
We are now in a position to evaluate a variety of trigonometric limits.
Example 1.13
2
2
2
2
Differential Calculus
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