Limits, Continuity & Differentiability
General
Grade 12
Question:
<p>Let a be a positive real number. Let f : R →R and g : (a, ∞) →R be defined by
f(x) = sin
πx
12
,
g(x) = 2 loge(√x −√a)
loge(e
√x −e
√a) .
Then the value of limx→a+ f(g(x)) is</p>
Step-by-Step Solution
Key Concept: General
<p>Put</p> t = \sqrt{x} -\sqrt{a.} Then t \to 0+ as x \to a+, and g(x) = 2 ln t ln e \sqrt{a}+t -e \sqrt{a}. Now e \sqrt{a}+t -e \sqrt{a} = e \sqrt{a}(et -1), so ln e \sqrt{a}+t -e \sqrt{a} = \sqrt{a} + ln(et -1). As t \to 0+, we have et -1 ∼t, hence ln(et -1) = ln t + o(1). Therefore, g(x) = 2 ln t ln t + \sqrt{a} + o(1) \to 2. Thus, lim x\to a+ f(g(x)) = f(2) = sin 2\pi 12 = sin \pi 6 = 1<p><strong>2</strong>: </p> Shortcut / Fast View When logarithms involve a vanishing expression, isolate the dominant ln t term and ignore bounded constants in comparison.
Correct Answer: (0.50)