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QUESTION: 1

The value of the limit

Solution:

QUESTION: 2

The value of the limit

Solution:

lim (x^{2}-9)/x^{3}+9x-6x^{2}

Lim x=0

lim[(0)^{2} - 9]/ (0)^{3} + 9(0) - 6(0)^{2}

-9/0 (which is not defined)

QUESTION: 3

Solution:

QUESTION: 4

Solution:

√x+1−1 / x

=(√x+1−1)(√x+1+1) / x(√x+1+1)

=x+1−1 / x(√x+1+1)

= 1/√x+1+1

So limx→0 √x+1−1/x

≡ limx→0 1/√(x+1)+1

= ½

QUESTION: 5

The value of the limit

Solution:

lt x->-2 (1/x + 1/2) / x+2

lt x->-2 (2+x)/2x(x+2)

lt x->-2 (1/2x)

= 1/2(-2) = -¼

QUESTION: 6

The value of the limit

Solution:

lim x → 2 [(2^{x} - 4) (2^{x} - 2) ((√2^{x}) + 2)]/[(√2^{x} - 2) (√2x - 2)]

= lim x → 2 [(2x - 4) (2 *4)]/(√2x - 2)

= 2 lim(x → 2) [(√2x)2 - (2)2]/(√2x - 2)

= 2 lim(x → 2) [(√2x)2 + 2) (√2x)2 - 2)]/(√2x - 2)

= 8

QUESTION: 7

Solution:

QUESTION: 8

The value of the limit

Solution:

QUESTION: 9

Solution:

QUESTION: 10

Solution:

limx→1x^{9}−1/x^{5}−1 has indeterminate initial form (it has form 0/0). Therefore 1is a zero and x−1 a factor of both the numerator and the denominator.

limx→1 x^{10}−1/x^{5}−1

= limx→1[(x−1)(x^{8}+x^{7}+x^{4}….⋅+x+1)]/(x−1)(x^{5}+x^{6}+x^{4}+⋅⋅⋅+x+1)

= (1+1+1+1+.....upto 10 terms)/(1+1+1+⋅⋅⋅upto 5 terms)

= 10/5 = 2

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