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Exam base material. Best fundamental topic. Top 5 exam base example.
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✓ The following are indeterminate forms which we will study:
0
0
∞
If lim
x → a
f
x
g
x
leads to the indeterminate form
or
then
lim
x → a
f
x
g
x
= lim
x → a
f
′
x
g
′
( x
, provided the later limit exists.
✓ Procedure to find the limit using L’ Hospital’s rule:
(1). Differentiate numerator and denominator separately and apply the limit.
(𝟐). If it again reduces to indeterminate form
or
then again
differentiate numerator and denominator separately and apply the limit.
. Continue this process till we get finite or infinite value of the limit.
✓ Remark: lim
x → 0
log x = −∞ & lim
x → ∞
log x = ∞.
Evaluate lim
x → 0
e
x
− 1 − x
x
2
Evaluate lim
x →
π
2
2x − π
cos x
Evaluate the examples: (𝟏). lim
x → 0
x − tan x
x
3
) (𝟐). lim
x → 0
sin x − x +
x
3
x
5
Evaluate lim
x → 0
2x − x cos x − sin x
2 x
3
Evaluate lim
x → 0
sin x
2
− x
2
x
2
( sin x
2
Evaluate the examples: (𝟏). lim
x → 0
tan x − x
sin x − x
) (𝟐). lim
x → 0
tan x − sin x
sin x
3
(𝟑). lim
x → 0
x
cos x − 1
sin x − x
) (𝟒). lim
x → 0
1 + x − 2 − x
2 sin
2
x
Evaluate lim
x → 0
e
x
−x
− x
2
sin x
2
− x
2
Evaluate lim
x → 0
e
x
− e
sin x
x − sin x
Evaluate lim
x →
1
2
cos
2
πx
e
2x
− 2ex
𝟐
Evaluate lim
x → y
x
y
− y
x
x
x
− y
y
Evaluate lim
x → 0
ln cos √
x
x
Evaluate lim
x → 0
xe
x
− log
1 + x
x
2
In this case we write f
x
∙ g
x
as
f
x
g
x
or
g
x
f
x
which leads to the form
or
, where L’ Hospital’s rule is applicable.
. Transform f
x
∙ g
x
into
f(x)
g
x
or
g(x)
f
x
i. e. transform 0 × ∞ into
or
(𝟐). Remember: Don’t put the logarithm function in the denominator in step (1).
. Apply L’ Hospital’s rule to find the value of given limit.
Evaluate lim
x → ∞
{ (a
1
x − 1 ) x }.
Evaluate lim
x → 1
1 − x
tan (
πx
Evaluate lim
x → ∞
x + 1 − √
x) log (
x
H 4 Evaluate lim
x → 0
{ (sin x) (ln x) }.
T 5 Evaluate the given examples:
(𝟏). lim
x → a
{ ln ( 2 −
x
a
) cot
x − a
} (𝟐). lim
x →
1
2
{ ln (
− x) cot (x −
In this case, we write f
x
− g
x
g
x
f
x
f
x
g
x
which leads to the form
or
where, L’ Hospital’s rule is applicable.
. Take L. C. M. in f
x
− g(x) which will transform ∞ − ∞ into
or
(𝟐). Apply L’ Hospital’s rule to find the value of given limit.
Evaluate lim
x → 0
sin x
x
H 2 Evaluate lim
x → 0
( cosec x − cot x).
State L
′
Hospital
′
s Rule. Use it to evaluate lim
x → 0
x
2
sin x
2
Evaluate lim
x → 0
x
2
cot x
2
Use L
′
Hospital
′
s rule to find the limit of lim
x → 1
x
x − 1
log x
Evaluate lim
x → 0
x
e
x
Evaluate lim
x → 2
x − 2
log(x − 1 )
C 3 Evaluate lim
x → 0
cos x
cot x
T 4 Evaluate given examples: (𝟏). lim
x → 1
x − 1
x− 1
(𝟐). lim
x →
π
2
sin x
tan x
(𝟑). lim
x →
π
2
{ (tan x)
cos x
Evaluate lim
x → 0
tan x
x
1
x
2
𝟏/𝟑
Evaluate lim
x → 0
a
x
x
x
1
3x
𝟏/𝟗
H 7 Evaluate the given examples:
(𝟏). lim
x → 0
x
x
x
x
1
x
(𝟐). lim
x → 0
a
x
x
x
x
1
x
𝟏/𝟒
𝟏/𝟒
Evaluate lim
x → 0
e
x
2x
3x
1
x
𝟐
Evaluate lim
x → 0
(a
x
1
x .
Evaluate lim
x → 1
2 − x
(tan
πx
2
)
𝟐/𝛑
Evaluate lim
x →
π
2
cos x
(
π
2
− x)
Evaluate the examples: (𝟏). lim
x →
π
2
cosec x
(tan
2
x)
} (𝟐). lim
x → 0
(cos √
x )
1
x
Evaluate lim
x → 0
x
(
1
ln(e
x
− 1 )
)
Evaluate lim
x → 1
( 1 − x
2
(
1
log( 1 −x)
)
✓ If the limit of integral is infinite (one or both) and/or integrand function is discontinuous
for some value(s) on the interval of given integral then such integral is known as an
improper integral.
✓ Types of Improper Integrals:
(1). Improper integral of first kind.
(2). Improper integral of second kind.
(3). Improper integral of third kind (combination of 1
st
nd
kind).
For ∫ f
x
dx
b
a
either a or b or both (a and b) are infinite, then such integral is known as an
improper integral of first kind.
✓ Sub-types of improper integrals of first kind are as follows:
(1). If f is continuous on [a, ∞) then
∫ f(x)dx = lim
b → ∞
∫ f(x)dx
b
a
∞
a
(2). If f is continuous on (−∞, b] then
Evaluate ∫
1 + x
2
dx
∞
−∞
Evaluate
x
1 + x
2
2
dx.
∞
−∞
Evaluate ∫ e
x
dx
∞
−∞
Evaluate ∫
e
x
−x
dx
∞
−∞
For ∫ f(x)dx
b
a
, f(x) become discontinuous (infinite) at x = a or x = b or at finite
number of points in the interval
a, b
, then such an integral is known as improper
integral of second kind.
✓ Sub-types of improper integrals of second kind are as follows:
(1). If x = a is the point of discontinuity for f(x) then the integral is defined as
∫ f
x
dx = lim
t → a
∫ f
x
dx
b
t
b
a
(2). If x = b is the point of discontinuity for f(x) then the integral is defined as
∫ f
x
dx = lim
t → b
∫ f
x
dx
t
a
b
a
(3). If x = c is the point of discontinuity for f(x) then the integral is defined as
∫ f
x
dx = ∫ f
x
dx + ∫ f
x
dx
b
c
c
a
lim
t → c
∫ f(x)dx
t
a
b
a
t → c
∫ f
x
dx
b
t
Evaluate the integrals ∫
x − 2
dx
5
2
and ∫
(a − x)
2
dx
b
a
Evaluate the integrals ∫
x
2
dx
1
0
and ∫
x
2 / 3
dx
1
− 1
Answer: ∞ & 6.
Evaluate
sec x dx
π
2
0
Evaluate
3 − x
dx.
3
0
Can we solve the integral ∫
x − 2
2
dx
5
0
directly? Give the reason.
Evaluate ∫
x − 1
2 / 3
dx.
3
0
𝟏/𝟑
𝟏/𝟑
𝟏/𝟑
Evaluate ∫
√a
2
− x
2
dx
a
−a
Check the convergence of ∫
x
dx
∞
1
and ∫
x
√ 2
dx
∞
1
Check the convergence of ∫
dx
1 + x
2
1 + tan
− 1
x
∞
0
Check the convergence of ∫
x
p
dx
∞
1
Check the convergence of ∫
sin
2
x
x
2
dx
∞
1
and ∫
sin
3
x
x
3
dx
∞
1
Check the convergence of ∫
log x
dx
∞
2
Check the convergence of ∫
x
1 + x
4
dx
∞
2
and ∫
x ( 1 + x)
2
dx
∞
1
Check the convergence of ∫ e
−x
2
dx
∞
1
Investigate the convergence of ∫
5x
1 + x
2
)
3
dx.
∞
5
Prove that ∫
e
x
dx
∞
1
is convergent.
Check the convergence of ∫
3x + 5
x
4
dx
∞
4
Investigate the convergence of
x
10
1 + x
5
1 + x
27
dx.
∞
0
Check the convergence of ∫
x
2
dx.
5
0
Determine whether the integral ∫
dx
x − 1
3
0
converges or diverges.
Determine
(x − 2 )
2
dx
2
0
and is it convergent or divergent?
Check the convergence of
dx
√ 1 − x
2
1
0
C 5 Define improper integral of both the kinds and check the convergence of
dx
√ 9 − x
2
3
0
Check the convergence of ∫
cos 3x
x
5 / 2
dx
3
0