Problem
Let f: (0, 1) → R be the function defined as f(x) = √n if x ∈ [1/{n + 1} , 1/n) where n ∈ N. Let g: (0, 1) → R be a function such that \( \int_{x^2}^{x}\sqrt{\frac{1-t}{t}}dt \) < g(x) < 2√x for all x ∈ (0, 1). Then \( \lim_{x \to 0}\) f(x)g(x)
A does NOT exist
B is equal to 1
C is equal to 2
D is equal to 3
Solution
Given the inequality involving g(x), we can guess that the problem involves using the Squeeze Theorem (also known as the Sandwich Theorem).
To proceed, we’ll try to establish a similar inequality for f(x), and then for f(x).g(x). If things go well, we’ll get the same limit for the upper and lower bounds of f(x).g(x).
For the inequality, let’s start with the definition of f(x). We have f(x) = √n if x ∈ [1/{n + 1} , 1/n).
Now, since x ∈ [1/{n + 1} , 1/n), 1/x ∈ [n , n + 1). And therefore, n ∈ (1/x – 1 , 1/x], which means √n ∈ (√(1/x – 1) , 1/√x].
Since √n equals f(x), we get √(1/x – 1) < f(x) ≤ 1/√x. Just what we needed.
Now, since f(x) and g(x) are both positive, we can ‘multiply’ the two inequalities to get:
\( \sqrt{\frac{1}{√x}-1}\int_{x^2}^{x}\sqrt{\frac{1-t}{t}}dt \) < f(x)g(x) < 2√x.1/√x
Now, let’s apply the limit x → 0 to each function.
\(\lim_{x \to 0}\sqrt{\frac{1}{x}-1}\int_{x^2}^{x}\sqrt{\frac{1-t}{t}}dt \) < \(\lim_{x \to 0}\) f(x)g(x) < \(\lim_{x \to 0}\) 2√x.1/√x
The upper bound simplifies to 2, and its limit is also 2. For the lower bound, we’ll need to do some work.
If we rearrange things a bit, the lower bound looks like this \(\lim_{x \to 0} \sqrt{1-x}\frac{\int_{x^2}^{x}\sqrt{\frac{1-t}{t}}dt}{\sqrt{x}} \)
Ignoring the \(\sqrt{1-x}\) (whose limit is 1), the rest of the function is of the 0/0 form, and can be evaluated using the L’Hôpital’s rule. It’s going to be a mess.
We’ll get \(\lim_{x \to 0} \frac{\sqrt{\frac{1-x}{x}} – \sqrt{\frac{1-x^2}{x^2}}.2x}{\frac{1}{2√x}} \), which simplifies to \(\lim_{x \to 0} 2 \sqrt{1-x} – 4\sqrt{x}\sqrt{1-x^2} \).
This limit is straightforward, and comes out to be 2.
Since both upper bound and lower bound limits equal 2, using the Sandwich theorem, the value of \(\lim_{x \to 0}\) f(x)g(x) is also 2.
Therefore, the correct option is C.
Comments
A well designed problem on limits. Getting to that inequality was the main step. Post that, it was fairly straightforward. It’s possible to get confused since the limit was asked as x → 0 and not x → 0⁺. As the left land limit is undefined, a student could’ve answered A, even after putting so much effort to get 2 as the right hand limit. Nevertheless, a good problem – that’s what the JEE Advanced is all about.
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