If f(x)= 1/3[f(x+1) + 5/f(x-2)], f(x)>0 for all x element of R and lim x tends to infinity f(x) exists finitely then lim x tends to infinity f(x) =

To solve the equation \( f(x) = \frac{1}{3} \left[ f(x+1) + \frac{5}{f(x-2)} \right] \) and find the limit as \( x \) approaches infinity, we can start by analyzing the behavior of \( f(x) \) as \( x \) increases. Given that \( f(x) > 0 \) for all \( x \in \mathbb{R} \) and that the limit exists finitely, we can denote this limit as \( L \). Our goal is to determine the value of \( L \).

Setting Up the Limit

Assuming that \( \lim_{x \to \infty} f(x) = L \), we can substitute \( L \) into the original equation. As \( x \) approaches infinity, \( f(x+1) \) and \( f(x-2) \) will also approach \( L \). Thus, we can rewrite the equation in terms of \( L \):

Substituting the Limit

Substituting \( L \) into the equation gives us:

\( L = \frac{1}{3} \left[ L + \frac{5}{L} \right] \)

Solving for L

Now, we can multiply both sides by 3 to eliminate the fraction:

\( 3L = L + \frac{5}{L} \)

Next, we can rearrange this equation:

\( 3L - L = \frac{5}{L} \)

which simplifies to:

\( 2L = \frac{5}{L} \)

Multiplying Through by L

To eliminate the fraction, we can multiply both sides by \( L \) (noting that \( L > 0 \)):

\( 2L^2 = 5 \)

Now, we can solve for \( L^2 \):

\( L^2 = \frac{5}{2} \)

Taking the square root of both sides gives us:

\( L = \sqrt{\frac{5}{2}} \)

Final Result

Since \( f(x) > 0 \) for all \( x \), we discard the negative root. Therefore, the limit as \( x \) approaches infinity is:

\( \lim_{x \to \infty} f(x) = \sqrt{\frac{5}{2}} \)

In summary, the limit of the function \( f(x) \) as \( x \) approaches infinity is \( \sqrt{\frac{5}{2}} \), which is approximately 1.581. This solution demonstrates how we can manipulate the functional equation to find the limiting behavior of the function at infinity.