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The graph of y = f(x) has a non-vertical tangent at the point (2, 0). Evaluate: lim(x->2) [ln(1 + 5f(x))] / [e^(2f(x)) - f(x) - 1].

1. 1
2. 5
3. 3
4. None of these

Correct answer: 5

Solution

As x->2, f(x)->f(2)=0 (since graph passes through (2,0)). Let t = f(x)->0. Numerator: ln(1+5t) ≈ 5t for small t. Denominator: e^(2t) - t - 1 ≈ (1 + 2t + 2t² +...) - t - 1 = t + 2t² +... ≈ t for small t. Limit = 5t/t = 5.

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