Why does the inequality $f(x) \le f(2x) \le f(3x)$ hold? Can't the function wiggle?

The problem defines $f(x)$ as an *increasing* function. This means for any $a < b$, $f(a) \le f(b)$. Since $x < 2x < 3x$ for any positive $x$, the function must be at a higher or equal level at $3x$ compared to $x$.

JEE Main 2010

LEVELJEE Main

Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $f : R \to R$ be a positive increasing function with $lim_{x \to \infty} \frac{f ( 3 x )}{f ( x )} = 1$ . Then $lim_{x \to \infty} \frac{f ( 2 x )}{f ( x )} =$

Q: Wait, why can we just divide by $f(x)$? What if $f(x)$ is zero?

Great catch! The problem explicitly states $f(x)$ is a positive function, meaning $f(x) > 0$ for all $x$. This guarantees we are never dividing by zero, keeping our inequality safe and sound.

Q: Can I use L'Hopital's Rule here?

L'Hopital's Rule requires us to know the derivative of $f(x)$, but we don't even know the function's explicit form! This problem is designed to test your conceptual understanding of limits and inequalities, not just mechanical differentiation.

Select Answer:

Visualized Solution

Understanding the Function

Given: $f (x)$ is a positive and increasing function.
Domain: $x > 0$ (since we are taking $x \to \infty$ ).

Ordering the Inputs

For any positive number $x$ , we can establish a clear order.
$x < 2 x < 3 x$

Applying the Increasing Property

Since $f (x)$ is an increasing function, larger inputs yield larger outputs.
Therefore: $f (x) \leq f (2 x) \leq f (3 x)$

Creating the Ratio

We need to find the limit of the ratio $\frac{f ( 2 x )}{f ( x )}$ .
Let's divide our entire inequality by $f (x)$ .

Safe Division

Divide by $f (x)$ : $\frac{f ( x )}{f ( x )} \leq \frac{f ( 2 x )}{f ( x )} \leq \frac{f ( 3 x )}{f ( x )}$
Note: Since $f (x) > 0$ , the inequality signs do not flip.

Simplifying the Inequality

The leftmost term simplifies: $\frac{f ( x )}{f ( x )} = 1$
The new inequality becomes: $1 \leq \frac{f ( 2 x )}{f ( x )} \leq \frac{f ( 3 x )}{f ( x )}$

Applying Limits

Take the limit as $x \to \infty$ across the inequality.
$lim_{x \to \infty} 1 \leq lim_{x \to \infty} \frac{f ( 2 x )}{f ( x )} \leq lim_{x \to \infty} \frac{f ( 3 x )}{f ( x )}$

Evaluating the Bounds

Left bound: $lim_{x \to \infty} 1 = 1$
Right bound: Given $lim_{x \to \infty} \frac{f ( 3 x )}{f ( x )} = 1$

The Sandwich Theorem

We now have: $1 \leq lim_{x \to \infty} \frac{f ( 2 x )}{f ( x )} \leq 1$
By the Sandwich Theorem (or Squeeze Theorem), the middle term is trapped.

Final Conclusion

The only possible value for the middle term is $1$ .
Final Answer: $lim_{x \to \infty} \frac{f ( 2 x )}{f ( x )} = 1$

00:00 / 00:00

The Sigma Insight: Evaluation of Limits & L'Hopital's Rule

Solution Diagram

Analyzing the Setup

The essence of an increasing function

f (x)

is that it never dips; as

x

increases,

f (x)

either stays the same or increases. We are given that

f (x) > 0

and is monotonically increasing.

We are provided with the critical limit:

x \to \infty lim \frac{f ( 3 x )}{f ( x )} = 1

Our objective is to determine the value of the limit:

x \to \infty lim \frac{f ( 2 x )}{f ( x )}

The Inequality Game

For any

x > 0

, we observe the relationship between the inputs:

x < 2 x < 3 x

. Because

f (x)

is a monotonically increasing function, it preserves this order.

This allows us to establish the following inequality:

f (x) \leq f (2 x) \leq f (3 x)

Since the problem guarantees

f (x) > 0

, we can divide the entire inequality by

f (x)

without reversing the inequality signs. This yields:

\frac{f ( x )}{f ( x )} \leq \frac{f ( 2 x )}{f ( x )} \leq \frac{f ( 3 x )}{f ( x )}

Simplifying the leftmost term, we arrive at:

1 \leq \frac{f ( 2 x )}{f ( x )} \leq \frac{f ( 3 x )}{f ( x )}

The Squeeze

Now, we evaluate the limit as

x \to \infty

for all three parts of the inequality:

x \to \infty lim 1 \leq x \to \infty lim \frac{f ( 2 x )}{f ( x )} \leq x \to \infty lim \frac{f ( 3 x )}{f ( x )}

The limit of the constant

1

is simply

1

. We are given that the rightmost limit is also

1

By the Squeeze Theorem (or Sandwich Theorem), since our unknown limit is trapped between

1

and

1

, it must converge to that same value.

Therefore, the final result is:

x \to \infty lim \frac{f ( 2 x )}{f ( x )} = 1

Common Traps & FAQs

Wait, why can we just divide by

f (x)

? What if

f (x)

is zero?

Great catch! The problem explicitly states

f (x)

is a positive function, meaning

f (x) > 0

for all

x

. This guarantees we are never dividing by zero, keeping our inequality safe and sound.

Why does the inequality

f (x) \leq f (2 x) \leq f (3 x)

hold? Can't the function wiggle?

The problem defines

f (x)

as an *increasing* function. This means for any

a < b

f (a) \leq f (b)

. Since

x < 2 x < 3 x

for any positive

x

, the function must be at a higher or equal level at

3 x

compared to

x

Can I use L'Hopital's Rule here?

L'Hopital's Rule requires us to know the derivative of

f (x)

, but we don't even know the function's explicit form! This problem is designed to test your conceptual understanding of limits and inequalities, not just mechanical differentiation.

Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $f : R \to R$ be a positive increasing function with $lim_{x \to \infty} \frac{f ( 3 x )}{f ( x )} = 1$ . Then $lim_{x \to \infty} \frac{f ( 2 x )}{f ( x )} =$

Select Answer:

Visualized Solution

Understanding the Function

Ordering the Inputs

Applying the Increasing Property

Creating the Ratio

Safe Division

Simplifying the Inequality

Applying Limits

Evaluating the Bounds

The Sandwich Theorem

Final Conclusion

The Sigma Insight: Evaluation of Limits & L'Hopital's Rule

Analyzing the Setup

The Inequality Game

The Squeeze

Similar Questions

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If $f (x)$ is differentiable and strictly increasing function, then the value of $lim_{x \to 0} \frac{f ( x ^{2} ) - f ( x )}{f ( x ) - f ( 0 )}$ is

Let $f : R \to R$ be such that $f (1) = 3$ and $f^{'} (1) = 6$ . Then $lim_{x \to 0} (\frac{f ( 1 + x )}{f ( 1 )})^{1/ x}$ equals

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For $x \in R, lim_{x \to \infty} (\frac{x - 3}{x + 2})^{x} =$

Select Answer:

Visualized Solution

Understanding the Function

Ordering the Inputs

Applying the Increasing Property

Creating the Ratio

Safe Division

Simplifying the Inequality

Applying Limits

Evaluating the Bounds

The Sandwich Theorem

Final Conclusion

The Sigma Insight: Evaluation of Limits & L'Hopital's Rule

Analyzing the Setup

The Inequality Game

The Squeeze

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