Limit

1 )

lim_x→2 [(4 – 8x + 5x² – x³) / (2x³ – 9x² + 12x – 4)] = ?

(1/3)

– (1/3)

– 3

View Solution

Answer :

in case of (0/0), lim_x→a [{f(x)} / {g(x)}] = lim_x→a [{f'(x)} / {g'(x)}]

∴ given is lim_x→2 [(0 – 8 + 10x – 3x²) / (6x² – 18x + 12)]

= lim_x→2 [(10 – 6x) / (12x – 18)]

= [{10 – (6 × 2)} / {(12 × 2) – 18}]

= [(– 2) / 6]

= – (1/3).

2 )

lim_x→3 [{√(x² + x + 3) – √(4x + 3)} / (x⁴ – 81)] = ?

[1 / (24√3)]

[1 / {72√(15)}]

[1 / (72√3)]

[1 / {24√(15)}]

View Solution

Answer :

lim_x→3 [{√(x² + x + 3) – √(4x + 3)} / (x⁴ – 81)]

= [{(lim_x→3[{(x² + x + 3)^(1/2) – (15)^(1/2)} / {(x² + x + 3) – 15}]) × lim_x→3(x + 4)

– (lim_x→3[{(4x + 3)^(1/2) – (15)^(1/2)} / {(4x + 3) – 15}]) × 4}

/ {lim_x→3[(x⁴ – 3⁴) / (x – 3)]}]

= [{(1/2)(15)^–(1/2)(3 + 4) – (4)(1/2)(15)^(1/2)} / (4 ∙ 3³)]

= [{(7/2) ∙ (1 / √15) – (2 / √15)} / (4 × 27)]

= [3 / (2√15)] × [1 / (4 × 27)]

= [1 / {72√(15)}].

3 )

lim_x→(π/3) [{sin[(π/3) – x]} / (2 cos x – 1)] is equal to

√3

(1/2)

(1 / √3)

(2 / √3)

View Solution

Answer :

lim_x→(π/3) [{sin[(π/3) – x]} / (2 cos x – 1)] [(0/0) form]

= lim_x→(π/3) [{– cos[(π/3) – x]} / (– 2 sin x)]

= [(cos 0) / {2 sin(π/3)}]

= [1 / {2 ∙ (√3 / 2)}]

= (1 / √3).

4 )

lim_x→0 [(sin 2x – tan 2x) / x³] = ?

-4

View Solution

Answer :

given: lim_x→0 [(sin 2x – tan 2x) / x³]

= lim_x→0 [{tan 2x(cos 2x – 1)} / x³]

= lim_x→0 ∙ [(tan 2x) / x] ∙ [(– 2 sin² x) / x²]

= lim_x→0 ∙ [(tan 2x) / 2x] ∙ (– 4) ∙ lim_x→0 [(sin x) / x]²

= (– 4)(1)(1)

= – 4.

5 )

lim_{x→–(π/4)} [{sin x ∙ cos(5π / 4) – cos(7π / 4) cos x} / (π + 4x)] = ?

– (1/3)

(35/4)

– (1/4)

– (1/35)

View Solution

Answer :

lim_{x→–(π/4)} [{sin x ∙ cos(5π / 4) – cos(7π / 4) ∙ cos x} / (π + 4x)]

= lim_{x→–(π/4)} [{sin x ∙ [– (1 / √2)] – (1 / √2)cos x} / (π + 4x)]

= lim_{x→–(π/4)} [{– sin x ∙ (1 / √2) – (1 / √2)cos x} / {4[(π/4) + x]}]

= (– 1) lim_{x→–(π/4)} [{sin[(π/4) + x]} / {4[(π/4) + x]}]

= [– (1/4)] lim_{x→–(π/4)} [{sin[(π/4) + x]} / {(π/4) + x}]

= – (1/4).

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