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QUESTION: 1

(k is a positive integer)

Solution:

Using – L- Hospital rule

QUESTION: 2

Solution:

On rationalizing

QUESTION: 3

Solution:

QUESTION: 4

Solution:

Given limit is

QUESTION: 5

Solution:

By L- Hospital rule

QUESTION: 6

Solution:

QUESTION: 7

Solution:

Divide numerator and denominator by x^{2}.

QUESTION: 8

Solution:

Using L hospital rule

lim x-->a d/dx[(x^{⅝} - a^{^⅝})/(x^{⅓} - a^{⅓})

lim x-->a [5/8(x^{-⅜})]/1/3(x^{-⅔})

lim x-->a 15/8 [x^{⅔}]/[x^{⅜}]

lim x-->a 15/8 [x^{⅔-⅜}]

Limx-->a 15/8 [x^{7/24}]

QUESTION: 9

Solution:

Using L-Hospital rule

QUESTION: 10

Solution:

Using L-Hospital rule

QUESTION: 11

Solution:

Using L- Hospital Rule

QUESTION: 12

Solution:

lim(x→0) sinx sin(π/3+x) sin(π/3−x)x

= lim(x→0) (sinx/x) sin(π/3+x) sin(π/3−x)

= 1 (sin(π/3) (sin(π/3)

= 1(√3/2)(√3/2)

= 3/4

QUESTION: 13

Solution:

QUESTION: 14

Solution:

Using L-Hospital rule

QUESTION: 15

Solution:

Use L- Hospital Rule

QUESTION: 16

Solution:

Given limit

QUESTION: 17

If = e^{2} then

Solution:

Given limit is

QUESTION: 18

Solution:

QUESTION: 19

Solution:

(1.2.3.....n)^{1/n}

QUESTION: 20

If [x] denotes the greatest integer less than or equal to x then

Solution:

By using Sandwitch theorem

QUESTION: 21

Let f : R → R be a positive increasing function with

Solution:

[x] can be written as X+{x} --> {x} is fractional part .. between 0 and 1

therefore ,

lim n→∞ {[x]+[2x]+[3x]+.........+[nx]}/n^{2}

is = lim n→∞ {x+2x+3x+.........+nx + {x}+{2x} ...+{nx}}/n^{2}

= {x(1+2+3+..+n) + {x}+...{nx}}/n^{2}

= x(n(n+1))/2n2 + ({x}+...{nx})/n^{2}

Since {} is only b/w 0 and 1 , the second operand becomes 0 as n tends to ∞

on solving the first part , u get (n^{2}x +nx)/2n^{2} = x/2 + x/2n

x/2n becomes 0 as x tends to infinity.

therefore the answer is x/2.

QUESTION: 22

Solution:

Using L- hospital rule

QUESTION: 23

Solution:

Take common higher power of x in both numerator and denominator

QUESTION: 24

Solution:

limx→0 [(1−cos2x)(3+cosx)]/xtan4x

= limx→0(1-(1-2sin^2x))(3+cosx)/xtan4x

= limx→0 (2sin/xx)(sinx)(3+cosx)/tan4x

= lim x→0 (2sinx/x)(sinx)(3+cosx)/(4x(tan4x/4x))

= limx→0(2sinxx)(sinx)3+cosx4xtan4x4x

We know, limx→0 sinx/x=1 and limx→0 tanx/x=1

∴ our expression becomes,

= 2/4 limx→0 (3+4cosx)

= 2/4limx→0 (3+4cosx)

As cos0=1

∴ our expression becomes, =2/4(3+1) = 2

QUESTION: 25

Solution:

QUESTION: 26

Solution:

QUESTION: 27

Solution:

Using standard formulae

QUESTION: 28

Solution:

1^{∞} form

QUESTION: 29

Solution:

e^{6-1} = e^{5}

QUESTION: 30

Solution:

Use L- Hospital rule

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