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Limits — JEE Maths practice question

JEE Maths question with a full step-by-step solution.

Question
f(x)=13(f(x+1)+5f(x+2))f(x)=\dfrac{1}{3}\left(f(x+1)+\dfrac{5}{f(x+2)}\right) and f(x)>0,  ∀x∈Rf(x)>0,\;\forall x\in\mathbb{R}, then lim⁡x→∞f(x)\displaystyle\lim_{x\to\infty}f(x) is
A00
B25\sqrt{\dfrac{2}{5}}
C52\sqrt{\dfrac{5}{2}}correct
D∞\infty
Solution
Step 1: Let lim⁡x→∞f(x)=K\displaystyle\lim_{x\to\infty}f(x)=K. Then lim⁡x→∞f(x+1)=K\displaystyle\lim_{x\to\infty}f(x+1)=K and lim⁡x→∞f(x+2)=K\displaystyle\lim_{x\to\infty}f(x+2)=K. Step 2: Take the limit of the given relation:
K=13(K+5K)K=\dfrac{1}{3}\left(K+\dfrac{5}{K}\right)
 Step 3: Multiply both sides by 3K3K:
3K2=K2+5  ⇒  2K2=5  ⇒  K2=523K^{2}=K^{2}+5\;\Rightarrow\;2K^{2}=5\;\Rightarrow\;K^{2}=\dfrac{5}{2}
 Step 4: Since f(x)>0f(x)>0, K>0K>0:
K=52K=\sqrt{\dfrac{5}{2}}
 Correct answer: (3) 52\sqrt{\dfrac{5}{2}}
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