The function f:R/\ 0 \ \to R given by f( x ) = 1 \over x - 2 \over e^2x - 1 can be made continuous at x = 0 by defining f (0) as

Limits Continuity and Differentiability PYQ — JEE Main 2007 · Shift 0 · Q40 with Solution

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Correct answer: B

To make $f$ continuous at $x=0$ , we must define $f(0)=\lim_{x\to 0}\left(\frac{1}{x}-\frac{2}{e^{2x}-1}\right).$ So we need to evaluate this limit.
Rewrite the expression using a common idea: $\frac{2}{e^{2x}-1}=\frac{1}{x}\cdot \frac{2x}{e^{2x}-1}.$ Hence, $f(x)=\frac{1}{x}\left(1-\frac{2x}{e^{2x}-1}\right).$ This is a $0/0$ type limit as $x\to 0$ .
Use the expansion of $e^{2x}$ near $x=0$ : $e^{2x}=1+2x+2x^2+\frac{4}{3}x^3+\cdots$ Therefore, $e^{2x}-1=2x+2x^2+\frac{4}{3}x^3+\cdots$ So,

=\frac{1}{x}\cdot \frac{1}{1+x+\frac{2}{3}x^2+\cdots}.$$ Now use $$\frac{1}{1+u}=1-u+u^2-\cdots$$ with $u=x+\frac{2}{3}x^2+\cdots$. Thus, $$\frac{1}{1+x+\frac{2}{3}x^2+\cdots}=1-x+\cdots$$ Hence, $$\frac{2}{e^{2x}-1}=\frac{1}{x}(1-x+\cdots)=\frac{1}{x}-1+\cdots$$ 4. Substitute into $f(x)$: $$f(x)=\frac{1}{x}-\left(\frac{1}{x}-1+\cdots\right)=1+\cdots$$ Therefore, $$\lim_{x\to 0}f(x)=1.$$ 5. So to make $f$ continuous at $x=0$, define $$f(0)=1.$$ 6. Checking options: - A: $0$ ❌ - B: $1$ ✅ - C: $2$ ❌ - D: $-1$ ❌ Therefore, the correct option is **B**.

View written solutionFree

Correct answer: B

To make $f$ continuous at $x=0$ , we must define $f(0)=\lim_{x\to 0}\left(\frac{1}{x}-\frac{2}{e^{2x}-1}\right).$ So we need to evaluate this limit.
Rewrite the expression using a common idea: $\frac{2}{e^{2x}-1}=\frac{1}{x}\cdot \frac{2x}{e^{2x}-1}.$ Hence, $f(x)=\frac{1}{x}\left(1-\frac{2x}{e^{2x}-1}\right).$ This is a $0/0$ type limit as $x\to 0$ .
Use the expansion of $e^{2x}$ near $x=0$ : $e^{2x}=1+2x+2x^2+\frac{4}{3}x^3+\cdots$ Therefore, $e^{2x}-1=2x+2x^2+\frac{4}{3}x^3+\cdots$ So,

=\frac{1}{x}\cdot \frac{1}{1+x+\frac{2}{3}x^2+\cdots}.$$ Now use $$\frac{1}{1+u}=1-u+u^2-\cdots$$ with $u=x+\frac{2}{3}x^2+\cdots$. Thus, $$\frac{1}{1+x+\frac{2}{3}x^2+\cdots}=1-x+\cdots$$ Hence, $$\frac{2}{e^{2x}-1}=\frac{1}{x}(1-x+\cdots)=\frac{1}{x}-1+\cdots$$ 4. Substitute into $f(x)$: $$f(x)=\frac{1}{x}-\left(\frac{1}{x}-1+\cdots\right)=1+\cdots$$ Therefore, $$\lim_{x\to 0}f(x)=1.$$ 5. So to make $f$ continuous at $x=0$, define $$f(0)=1.$$ 6. Checking options: - A: $0$ ❌ - B: $1$ ✅ - C: $2$ ❌ - D: $-1$ ❌ Therefore, the correct option is **B**.