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 0 f(x) exists finitely then lim x tends to 0 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 0, we need to analyze the behavior of the function \( f(x) \) under the given conditions. The fact that \( f(x) > 0 \) for all \( x \) in the real numbers and that the limit exists finitely as \( x \) approaches 0 will guide our exploration.

Setting Up the Equation

We start with the functional equation:

\( f(x) = \frac{1}{3} \left[ f(x+1) + \frac{5}{f(x-2)} \right] \)

This equation suggests a recursive relationship between the values of the function at different points. To find the limit as \( x \) approaches 0, we can assume that \( f(x) \) approaches some finite limit \( L \) as \( x \) approaches 0.

Assuming the Limit Exists

Let’s denote:

• \( L = \lim_{x \to 0} f(x) \)
• \( \lim_{x \to 0} f(x+1) = L \)
• \( \lim_{x \to 0} f(x-2) = L \)

Substituting these limits into the original 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:

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

Rearranging this equation leads to:

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

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

Next, we multiply both sides by \( L \) (noting that \( L > 0 \)):

\( 2L^2 = 5 \)

Dividing both sides by 2 gives:

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

Taking the square root of both sides, we find:

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

Final Result

Thus, the limit as \( x \) approaches 0 for \( f(x) \) is:

lim_x→0 f(x) = √(5/2)

This result indicates that as \( x \) gets closer to 0, the function \( f(x) \) stabilizes around the value \( \sqrt{\frac{5}{2}} \), which is approximately 1.58. This analysis shows how functional equations can be approached systematically to uncover limits and behaviors of functions.