Limits

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2 Answers

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Himanshu Giria ·2014-06-11 22:43:58

c

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Himanshu Giria is it c ??
Upvote·0· Reply ·2014-06-11 22:44:10
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1357

Manish Shankar ·2014-06-11 23:35:21

it becomes

[3f(0^-]-f(0)=3*3-4=5

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Swastik Haldar I don't understand the working inside the first G.I.F. why 3f(0-)? why not just 3f(0)?
Upvote·0· Reply ·2014-06-12 03:39:10
Manish Shankar because that number is less than zero but very close to zero as x=(0-)
Upvote·0· Reply ·2014-06-12 23:20:38
Himanshu Giria Sir but one thing I didnt understand is the how f(0-)=3
Upvote·0· Reply ·2014-06-13 00:25:52
Himanshu Giria I mean to say how are we solving the limit inside f() and how are we getting that to be 0-
Upvote·0· Reply ·2014-06-13 00:28:50
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Basic Integral Calculus Operators Symbols Matrix Cases

√ √ || {} [] () → ~ ^ x x x → → ln log lg lim lim cos sin tan cot arcsin arccos arctan

∫ ∫ ∫ ∫ ∫ ∬ ∬ ∬ ∬ ∬ ∭ ∭ ∭ ∭ ∭ ⨌ ⨌ ⨌ ⨌ ⨌

∑ ∑ ∑ ∑ ∑ ∮ ∮ ∮ ∮ ∮ ∏ ∏ ∏ ∏ ∏ ∐ ∐ ∐ ∐ ∐

+ - × ÷ ± ∓ = ∪ ∩ ≠ ∼ ≈ ∅ ∀ ∃ ∈ ∋ ∝ ≤ ≥ ≃ ≅ ≡ < > ≪ ≫ ⊂ ∉ ⊃ ⊆ ⊇ ⇒ ⇐ ⇔ ⇒ ⇐ ⇔ ⇑ ⇓ ⇕ → ← ↔ → ← ↔ ↑ ↓ ↕ ↖ ↗ ↘ ↙ ↪ ↩ ⇀ ↼ ⇁ ↽

° ∂ ∞ α β γ δ θ λ μ ν ξ π ρ σ φ ω ε ƒ κ τ υ Γ Δ Λ Σ Π Ω