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QUESTION: 1

Solution:

lim (x → 0) [((1-3x)+5x)/(1-3x)]^{1/x}

lim (x → 0) [1 + 5x/(1-3x)]^{1/x}

= e^{lim(x → 0) (1 + 5x/(1-3x) - 1) * (1/x)}

= e^{lim(x → 0) (5x/(1-3x)) * (1/x)}

= e^{lim(x → 0) (5x/(1-3x))}

= e^{5}

QUESTION: 2

Solution:

QUESTION: 3

Solution:

lim(x→0) [log10 + log1/10]/x

= [log10 + log10]/0

= 0/0 form

lim(x→0) [(1/(x+1/10) * 1]/1

lim(x→0) [(1/(0+1/10) * 1]/1

= 1/(1/10) => 10

QUESTION: 4

Solution:

QUESTION: 5

Solution:

lim(x → 1) (log2 2x)1/^{log}_{2}^{x}

= lim(x →1) (log22 + log2x)1/^{log}_{2}^{x}

As we know that {log ab = log a + log b}

lim(x → 1) {1 + log_{2}x}1/^{log}_{2}^{x}

log_{2}x → 0

Put t = log_{2}x

lim(t → 0) {1 + t}^{1}/t

= e

QUESTION: 6

Solution:

QUESTION: 7

lim(x → 0) (tanx/x)^{(1/x^2)}

Solution:

lim(x → 0) (tanx/x)^{(1/x^2)}

= (1)∞

e^{lim(x → 0) (1/x2)(tanx/x - 1)}

= e^{lim(x → 0) ((tanx - x)/x3)} .....(1)

lim(x → 0) ((tanx - x)/x^{3})

(0/0) form, Apply L hospital rule

lim(x → 0) [sec^{2}x -1]/3x^{2}

lim(x → 0) [tan^{2}x/3x^{2}]

= 1/3 lim(x → 0) [tan^{2}x/x^{2}]

= 1/3 * 1

= e^{1/3}

QUESTION: 8

Solution:

QUESTION: 9

Solution:

lim(x → ∞) [(x-2)/(x+3)]^{2x}

lim(x → ∞) [(x-2)/(x+3)]^{2} * [(x-2)/(x+3)]^{x}

lim(x → a) [f(x) * g(x)] = lim(x → a) f(x) * lim(x → a) g(x)

lim(x → ∞) [(x-2)/(x+3)]^{2} * lim(x → ∞) [(x-2)/(x+3)]^{x}

Lets evaluate the limits of both the functions separately,

lim(x → ∞) [(x-2)/(x+3)]^{2}

= lim(x → ∞) [(x(1-2/x)/(1+3/x)]^{2}

lim(x → ∞) [(1-2/x)/(1+3/x)]^{2}

Apply infinity property,

= [(1-0)/(1-0)]^{2}

= 1

Now, lim(x → ∞) [(x-2)/(x+3)]^{2}

= lim(x → ∞) exln[(x-2)/(x+3)]

= lim(x → ∞) ln[(x-2)/(x+3)]/(1/x)

Apply L hospital rule

lim(x → ∞) d/dx[ln(x-2)/(x+3)]/[d/dx(1/x)]

= lim(x → ∞) {1/[(x-2)/(x+3)] * d/dx[(x-2)/(x+3)]}/(-1/x2)

= lim(x → ∞) {(x+3)/(x-2)[(x+3)d/dx(x-2) - (x-2)d/dx(x+3)]/(x+3)^{2}}/(-1/x2)

= lim(x → ∞) {(x+3)/(x-2)[(x+3)-(x-2)]/(x+3)2}/(-1/x2)

= lim(x → ∞) {(x+3)/(x-2)[5/(x+3)^{2}]}/(-1/x2)

= lim(x → ∞) [-5x2/(x+3)(x-2)]

= lim(x → ∞) [-5x2/(x2 + x - 6)]

Again apply L hospital rule,

= lim(x → ∞) [d/dx(-5x2)/(d/dx(x2 + x - 6))]

= lim(x → ∞) [-10x/(2x + 1)]

Again applying L hospital rule,

= lim(x → ∞) [d/dx(-10x)/(d/dx(2x + 1))]

= lim(x → ∞) [-10/2]

= lim(x → ∞) [-5]

= -5

= lim(x → ∞) exln[(x-2)/(x+3)]^{2} = e(-5)2

= 1/e10

= lim(x → ∞) [(x-2)/(x+3)](x+3) = 1/e^{10 }(1)

= 1/e^{10} = e^{-10}

QUESTION: 10

If a,b,c,d are positive, then

Solution:

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