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Differential Calculus and Applications with sample problems

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Limits and functions 1. Theorem on Limits 2. Direct Substitution Method 3. Factorization Method 4. Rationalization Method 5. Infinity Method 6. L’Hospital’s Rule Asymptotes 1. Horizontal Asymptote 2. Vertical Asymptote 3. Inclined Asymptote Basic Differentiation 1. Derivatives o...

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  • December 3, 2022
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  • 2021/2022
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Differential Calculus


Topic Overview :



A .
limits and Functions
A. 1 . Theorems on limits

A. 2 . Direct substitution Method
A. 3 . Factorization Method
A. 4 .
Rationalization Method
A. 5 .
Infinity Method

A. 6 . L' Hospital's Rule

A.7 . Gille Sania 's Principle for limits

B. Asymptotes
B. l . Horizontal
Asymptotes
Vertical
B. 2 .

Asymptotes
B. 3 . Inclined Asymptotes
C .
Basic Differentiation

c. 1 . Derivatives of Functions at specific Values
c. 2 .
Implicit Differentiation
c. 3 .
Higher Derivatives

D . Basic Applications
D. I . Motion Problems

D. 2 .

Tangent & Normal lines

D. 3 .
Critical Points & points of Inflection

D. 4 .
Slope of a curve

E . Maxima and Minima Applications
F. Time Related Rates

G . Parametric Equation
H .
Curvature and Radius of curvature

I . Curvilinear Motion

J . Partial Differentiation and Applications

,A . LIMITS AND FUNCTIONS

A. l .
Theorems on limits
If f is
13 . a
polynomial :

"
Assume that lift tax ,
)
and ¥794 exist and that limtcx ) -
-
Ha)

any constant
c is .
Then ,
x →a



tim c =
c
1 ,

x→ a




lim x =a
The limit lim f- (x) -
-
M
2 .




x→ a X la
-




limlcfcxl) climflx) if and
only if the right hand limits and left hand
-
-
-
-


z .




x -
sa x→ a limits exist and are
equal to M :


limflx) limfcx) M
limlflxltglx))
=

limglx)
-




limfcx) I
-




¢
-
-
.




x → at
-




x→a xta x sa -
x sa-




s .
limltcx)g( ) ) x
= limfcx) .


limgcx) Suppose that f and
g are two functions such that
x la
Iim f- Cx) =L to and limglx )
-
x→ a x⇒a -
-
o


limflx) xta x→a


lim tht Ha
limglx) to then the limit f- Ix ) does not exist .



↳ = ,
lim
g Cx ) limg (x)
x 'a
g (x )
-


x sa
-


x sa-




x la
-




" "

7. limlflx) ) -
-
flimflx) ) ,
where n EN
x -
sa x→ a

"

8 .
lim x = an
x →a



limn f- Cx) limflx ) lim f- (x ) > o if n
q
= n
,
is
,




sa x→a even
x 'a -
x -




to . limllntlx)) -
-
lnflimflx ) ) ,
lim f- (x) > o

x →a x→ a x sa-




11 .
Squeeze Rule :


If ffx) Eg (x ) E h (x ) for all x in an
open
interval that contains a , except possibly at x=a
,



limtcx) =
limhlx) =L limgcx) =L
and , then
x -

sa x
-
ta x→a


12 . Composition Rule :


If f- (x) is continuous at x
-
- b and limglx) - b
x→a



then , limffglx) ) =
fllimgcx))
x sa-
x→a

, Sample Problems :




Example No 1 -
Example No 4 ,




Lim 721-52-1-6 him 3×3-2×+4
2- → I Ztl X -70 2 -
3×2-2×3
Solution : solution :



Lim z2t5zt6 lim (172+511)+6 him 3×3-2×+4
=


2- → I Zt I
2- → l Itt x -7N 2 -
3×2-2×3
=
6 ( answer) recall :



limflx) -
-
limfcx) =M
+
-




Example No 2 x→a x sa-

.




'
Lim x -
y

x -12 x' +2×-8 X =
999,999
say '

Solution : Lim 31999,999) -
21999,999)t4
2 3

Lim X
'
-
4 lim ( Xt 2) ( x -2) x -7999,999 2 -
31999,999) -
21999,999)
=


x' 1-2×-8 (Xt 4) ( X 2) 1.5
= -



x -12 X -72 -




lim xtz
=

X -72 Xt 4 X =
-999,999
say
lim 21-2 him 31-999,999/-21-999,999)t4
=
2
3
X -72 2+4 x -5999,999 2 -

31-999,999) -
21-999,999)
= -

1.5
=
f- } = ( answer )


: Lim 3×3-2×+4
( answer)
.



1.5
-
=



+ → a 2 3×2-2×3
Example No 3
-



,




Lim I
×
x→o


Solution :


recall :
Lim I )


x -70
×
limflx) -
-
limfcx) - M
t
-




x→a x sa-




say
X=
-

0.000001 ( left side of O )
^


Lim 1

X -7 0.000001 0.000001
- -




H it )
( ,
.




1,000,000
= -




-




V



X= 0.000001 ( right side of o )
say
I
Lim
X -70.000001 0.000001

= 1,000,000
#




since -1,000,000 ¥ 1,000,000 ,




- : LIMIT DOES NOT EXIST ( answer)

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