Let f : (0,1) → R be the function defined as f(x) = √n if x ∈ [\(\frac{1}{n+1},\frac{1}{n}\)] where n ∈ N. Let g : (0,1) → R be a function such that \(\int_{x^2}^{x}\sqrt{\frac{1-t}{t}}dt<g(x)<2\sqrt x\) for all x ∈ (0,1).
Then \(\lim_{x\rightarrow0}f(x)g(x)\)
Let f : (0,1) → R be the function defined as f(x) = √n if x ∈ [\(\frac{1}{n+1},\frac{1}{n}\)] where n ∈ N. Let g : (0,1) → R be a function such that \(\int_{x^2}^{x}\sqrt{\frac{1-t}{t}}dt<g(x)<2\sqrt x\) for all x ∈ (0,1).
Then \(\lim_{x\rightarrow0}f(x)g(x)\)
- does NOT exist
- is equal to 1
- is equal to 2
- is equal to 3
The Correct Option is C
Approach Solution - 1
\(f(x)=\sqrt{\frac{1}{x}-1}\)
\(\lim_{x\rightarrow 0^+}\int_{x^2}^{x}\sqrt{\frac{1-t}{t}}dt\sqrt{\frac{1}{x}-1}\times2\sqrt x\)
\(=\lim_{x\rightarrow 0^+}2\sqrt{x(\frac{1}{x}-{\frac{1}{x}})}=2\)
\(\lim_{x\rightarrow 0^+}2\sqrt {x(\frac{1}{x})}=2;(\frac{1}{x}\notin Z)\)
\(\lim_{x\rightarrow0^+}\int_{x^2}^{x}\sqrt{\frac{1-t}{t}}dt.\sqrt{\frac{1}{x}-\frac{1}{x}}=\int_{x^2}^{x}\frac{1-t}{t}dt\sqrt{1-x}(\frac{1}{x})\)
\(\lim_{x\rightarrow}2\sqrt{1-x}.4\sqrt{x}.\sqrt{1-x^2}=2\)
similarly for \(\frac{1}{x}\in Z \) is equal to 2.
Approach Solution -2
We need to solve a one-sided limit to find the answer, otherwise \(\lim\limits_{x→0^-}\) doesn't exist because it is not in the domain.
\(f(x)=\sqrt{(\frac{1}{x})-1}\) where (.) = least integer function
\(\lim\limits_{x→0^+}\int\limits_{x^2}^x\sqrt{\frac{1-t}{t}}dt.\sqrt{(\frac{1}{x})-1}\le \lim\limits_{x→0^+}f(x).g(x)\le\lim\limits_{x→o^+}\sqrt{(\frac{1}{x}-1)}\times2\sqrt{x}\)
So, \(\lim\limits_{x→0^+}\sqrt{(\frac{1}{x})-1}\times2\sqrt{x}=\lim\limits_{x→0^+}2\sqrt{x}\sqrt{[\frac{1}{x}]}\ \ (\frac{1}{x}∉ Z)\)
\(=\lim\limits_{x→0^+}2\sqrt{x\left(\frac{1}{x}-\left\{\frac{1}{x}\right\}\right)}=2\)
\(=\lim\limits_{x→0^+}2\sqrt{x(\frac{1}{2})}=2;(\frac{1}{x}∉Z)\)
\(\lim\limits_{x→0^+}\int\limits_{x^2}^{x}\sqrt{\frac{1-t}{t}}dt.\sqrt{\frac{1}{x}-\left\{\frac{1}{x}\right\}}=\frac{\int\limits_{x^2}^x\sqrt{\frac{1-t}{t}}dt.\sqrt{1-x\left\{\frac{1}{x}\right\}}}{\sqrt{x}}\)
\(\lim\limits_{x→0^+}\frac{\int\limits_{x^2}^x\sqrt{\frac{1-t}{t}dt}}{\sqrt{x}}=\lim\limits_{x→0^+}\frac{\sqrt{\frac{1-x}{x}}-2x\sqrt{\frac{1=x^2}{x^2}}}{\frac{1}{2\sqrt{x}}}\)
\(\lim\limits_{x→0^+}2\sqrt{1-x}-4\sqrt{x}.\sqrt{1-x^2}=2\)
In the same way, \(\frac{1}{4}\in Z\) is equal to 2.
So, the correct option is (C) : is equal to 2.
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Concepts Used:
Methods of Integration
Given below is the list of the different methods of integration that are useful in simplifying integration problems:
Integration by Parts:
If f(x) and g(x) are two functions and their product is to be integrated, then the formula to integrate f(x).g(x) using by parts method is:
∫f(x).g(x) dx = f(x) ∫g(x) dx − ∫(f′(x) [ ∫g(x) dx)]dx + C
Here f(x) is the first function and g(x) is the second function.
Method of Integration Using Partial Fractions:
The formula to integrate rational functions of the form f(x)/g(x) is:
∫[f(x)/g(x)]dx = ∫[p(x)/q(x)]dx + ∫[r(x)/s(x)]dx
where
f(x)/g(x) = p(x)/q(x) + r(x)/s(x) and
g(x) = q(x).s(x)
Integration by Substitution Method
Hence the formula for integration using the substitution method becomes:
∫g(f(x)) dx = ∫g(u)/h(u) du
Integration by Decomposition
Reverse Chain Rule
This method of integration is used when the integration is of the form ∫g'(f(x)) f'(x) dx. In this case, the integral is given by,
∫g'(f(x)) f'(x) dx = g(f(x)) + C
Integration Using Trigonometric Identities
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