Problem 46 - Single Variable Calculus

Question

Prove the inequality $$ \frac{x}{x+1}<\ln (1+x)0\).

Step-by-Step Solution

Answer

The inequality \(\frac{x}{x+1}<\ln (1+x) 0\) can be proven by analyzing it in two parts: the left inequality \(\frac{x}{x+1}<\ln (1+x)\) and the right inequality \(\ln (1+x) 0\) and that they are equal to 0 when \(x=0\). Thus, we can conclude that the given inequality is true for all \(x > 0\).

Step 1: Left Inequality Proving \(\frac{x}{x+1}<\ln (1+x)\)

To prove the left inequality \(\frac{x}{x+1}<\ln (1+x)\), let's consider a function \(f(x) = \ln(1+x) - \frac{x}{x+1}\). It is obviously defined for \(x>0\). Our goal is to show that \(f(x) > 0\) for all \(x > 0\).

Step 2: Find the derivative of f(x)

To analyze f(x), we will find its derivative with respect to x, \(f'(x)\). By applying the chain rule and quotient rule, we get: \(f'(x) = \frac{1}{1+x} - \frac{1(x+1) - x}{(x+1)^2} = \frac{1}{1+x} - \frac{1}{(x+1)^2}\)

Step 3: Simplifying the derivative

Now, let's simplify the derivative: \(f'(x) = \frac{(x+1) - 1}{(x+1)^2} = \frac{x}{(x+1)^2}\) Since \(x > 0\) and \((x+1)^2 > 0\), it's clear that the derivative \(f'(x) > 0\) for all \(x > 0\).

Step 4: Using the monotonic property

Since \(f'(x) > 0\), the function \(f(x)\) is monotonically increasing for \(x > 0\). Thus, we only need to check the limit as x approaches 0: \(\lim_{x \to 0} f(x) = \lim_{x \to 0} \left[ \ln(1+x) - \frac{x}{x+1} \right] = \ln(1) - \frac{0}{1} = 0\) Since \(f(x)\) is monotonically increasing and \(f(0)= 0\), it follows that \(f(x) > 0\) for all \(x > 0\). Thus, the left inequality \(\frac{x}{x+1}<\ln (1+x)\) is true for all \(x > 0\).

Step 5: Right Inequality Proving \(\ln (1+x) < x\)

To prove the right inequality \(\ln (1+x) < x\), let's consider a function \(g(x) = x - \ln (1+x)\). It is also defined for \(x>0\). Our goal is to show that \(g(x) > 0\) for all \(x > 0\).

Step 6: Find the derivative of g(x)

To analyze g(x), we will find its derivative with respect to x, \(g'(x)\). By applying the chain rule, we get: \(g'(x) = 1 - \frac{1}{1+x}\)

Step 7: Simplifying the derivative

Now, let's simplify the derivative: \(g'(x) = \frac{x}{x+1}\) Since \(x > 0\) and \(x+1 > 0\), it's clear that the derivative \(g'(x) > 0\) for all \(x > 0\).

Step 8: Using the monotonic property

Since \(g'(x) > 0\), the function \(g(x)\) is monotonically increasing for \(x > 0\). Thus, we only need to check the limit as x approaches 0: \(\lim_{x \to 0} g(x) = \lim_{x \to 0} \left[ x - \ln(1+x) \right] = 0 - \ln(1) = 0\) Since \(g(x)\) is monotonically increasing and \(g(0)= 0\), it follows that \(g(x) > 0\) for all \(x > 0\). Thus, the right inequality \(\ln (1+x) < x\) is true for all \(x > 0\).

Step 9: Combine the inequalities

Finally, combining the results from steps 4 and 8, we have proven that \(\frac{x}{x+1}<\ln (1+x) 0\).