This document provides guidance on computing limits and derivatives. It outlines proper techniques for evaluating limits involving infinity, indeterminate forms, and piecewise functions. Examples are given for using identities to simplify limits, implicit differentiation, and determining equations for asymptotes and tangent lines.

1. Department of Computer Science
Revision
March 30, 2015
1 Computing limits
1. a2
+ b2
= (a − b)(a + b).
2. For a = 0, if limx→x0
f(x) = 0
a
= 0. Do not write that the limit does not exist. The limit
exists and it is 0.
3. when the limit is in the form
0
0
, do not put it equal to 1. The same thing for
∞
∞
,
it is not 1.
4. To compute lim
x→0
√
x + 1 − 1
x
, we use the following identity a2
− b2
= (a − b)(a + b). This
will help us to get rid of the square root.
lim
x→0
√
x + 1 − 1
x
= lim
x→0
√
x + 1 − 1
x
·
√
x + 1 + 1
√
x + 1 + 1
= lim
x→0
√
x + 1
2
− 12
x
√
x + 1 + 1
= lim
x→0
x + 1 − 1
x
√
x + 1 + 1
= lim
x→0
1
√
x + 1 + 1
=
1
2
5. To compute limx→∞
x2
− x, do not use ∞ − ∞ = ∞. The correct way to compute the limit
is as follows.
lim
t→∞
x2
− x = lim
t→∞
x2
1 −
1
x
= lim
t→∞
x2
(1 − 0) because lim
t→∞
1
x
= 0
= ∞
6. The asymptotes are strongly related to the limits that involve infinity. So, to find the
equation of any asymptote, you have to compute the corresponding limits.

2. 7. To compute the limit for a function which is in piecewise form,
f(x) =
h−(x) if x ≤ a
h+(x) if x > a
do not forget to use the left-side limit
lim
x→a−
f(x) = lim
x→a−
h−(x)
as well as the right-side limit
lim
x→a+
f(x) = lim
x→a+
h+(x).
8. A function is differentiable at x = a if f (a) exists.
9. A differetiable function at x = a is a continuous function at x = a. But, a continuous
function at x = a is not neccessary differentiable at x = a.
10. Implicit differentiation use the chain rule. Let x2
+ xy2
= 2. implicit differentiation leads
to
d
dx
x2
+ xy2
=
d
dx
2
d
dx
x2
+
d
dx
xy2
= 0
2x + y2
+ x
d
dx
y2
= 0 product rule
2x + y2
+ x (2yy ) = 0 chain rule
y = −
2x + y2
2y
, where y = 0.
The equation of the tangent line at (1,1) is given by
y |(1,1) = −
3
2
y = −
3
2
(x − 1) + 1 point-slope formula