Limit

11 )

lim_x→π [{25 – √(626 + Cos x)} / (π – x)²] = ?

0.1

– 0.02

– 0.01

– 0.1

View Solution

Answer :

lim_x→π [{25 – √(626 + Cos x)} / (π – x)²]

∵ lim_x→a [{f(x)} / {g(x)}] = lim_x→a [{f'(x)} / {g'(x)}] .... [{f(x)} / {g(x)}] = (0/0)

∴ We can write,

lim_x→π [{0 – [1 / {2√(626 + Cos x)}](– Sin x)} / {2(π – x)(– 1)}]

= lim_x→π [(Sin x) / {2√(626 + Cos x)}] × [(– 1) / 2] × [1 / (π – x)]

lim_x→π [(– 1) / 4] [(Sin x) / {(π – x)√(626 + Cos x)}]

lim_x→π[(– 1) / 4] [(Cos x) / {(– 1)√(626 + Cos x) + (π – x)

∙ [(– Sin x) / {2√(626 + Cos x)}]}]

= [– (1/4)] [(– 1) / {(– 1) × 25 + 0}]

= [(– 1) / (100)]

= – 0.01.

12 )

lim_x→0 (3 / x³)sin(π² + 2x) – (3 / x³)sin(π² + x)– (1 / x³)sin(π² + 3x) + (1 / x³)sin[π(1 + π)] = ?

cos π^2

– cos π^2

– π

View Solution

Answer :

lim_x→0 (3 / x³)sin(π² + 2x) – (3 / x³)sin(π² + x) – (1 / x³)sin(π² + 3x)

+ (1 / x³)sin[π(1 + π)]

= (0/0) form

∴ on differentiation

lim_x→0 [{3 cos(π² + 2x)(2)} / (3x²)] – [{3 cos(π² + x)} / (3x²)]

– [{cos(π² + 3x)(3)} / (3x²)] + (0 / 3x²)

= lim_x→0 [{(– 2)sin(π² + 2x)(2)} / 2x] + [{sin(π² + x)} / 2x]

+ [{3 sin(π² + 3x)} / 2x]

= lim_x→0 [{(– 4)cos(π² + 2x)(2)} / 2] + [{cos(π² + x)} / 2]

+ [{9 cos(π² + 3x)} / 2]

= [{– 8 cos π² + cos π² + 9 cos π²) / 2]

= cos π².

13 )

lim_x→0 [{tan[(π/3) + 2x] – 2 tan[(π/3) + 2x] + tan(π/3)} / x] = ?

4√3

– 8√3

– 4√3

– 8

View Solution

Answer :

lim_x→0 [{tan[(π/3) + 2x] – 2 tan[(π/3) + 2x] + tan(π/3)} / x]

= lim_x→0 [{[{tan(π/3) + tan 2x} / {1 – tan(π/3) ∙ tan 2x}] – 2

[{tan(π/3) + tan 2x} / {1 – tan(π/3) ∙ tan 2x}] + √3} / x]

= lim_x→0 [{[(√3 + tan 2x) / (1 – √3 tan 2x)] – 2

[(√3 + tan 2x) / (1 – √3 tan 2x)] + √3} / x]

= lim_x→0 [(– √3 – tan 2x + √3 – 3 tan 2x) / {x(1 – √3 tan 2x)}]

= lim_x→0 [(– 4 tan 2x) / {x(1 – √3 tan 2x)}]

= – 8 × lim_x→0 [(tan 2x) / 2x] × lim_x→0 [1 / (1 – √3 tan 2x)]

= – 8 × 1 × lim_x→0 [(1 + √3 tan 2x) / (1 – 3 tan² 2x)]

= (– 8) × 1

= – 8.

14 )

lim_x→π (x – [x – 3] – [x – 3] = ? Where x ∈ (π – 0.1, π + 0.1) – {π}

– (π + 1)

π + 1

π – 1

View Solution

Answer :

∵ π = 3.14

∴ [x – 3] = [0.14] = 0

and

[3 – x] = [– 0.14] = – 1

∴ lim_x→π (x – [x – 3] – [3 – x]) = π – (0) – (– 1) = π + 1.

15 )

lim_θ→0 [{cos²(1 – cos²(1 – cos²(1…..cos² θ)))}/ {sin[{π[√(θ + 4)] – 2} / θ]}]

√2

– √2

View Solution

Answer :

lim_θ→0 [{cos²(1 – cos²(1 – cos²(1…..cos² θ)))} / {sin[{π[√(θ + 4)] – 2} / θ]}]

= lim_θ→0 [{cos²(sin²(sin²…..(sin² θ)))} / {sin[{π[√(θ + 4)] – 2} / θ]}]

= [{lim_θ→0 cos²(sin²(sin²…..(sin² θ)))}

/ {sin[π lim_θ→0 {θ / (θ[√(θ + 4) + 2])}]}]

= [(cos² θ) / {sin[π × (1/4)]}]

= [1 / (1 / √2)]

= √2

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