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Question
If \[f\left( x \right) = \frac{2x + 3\ \text{ sin }x}{3x + 2\ \text{ sin } x}, x \neq 0\] If f(x) is continuous at x = 0, then find f (0).
Solution
Given:
\[\lim_{x \to 0} f\left( x \right) = f\left( 0 \right)\]
\[ \Rightarrow \lim_{x \to 0} \frac{2x + 3\ sinx}{3x + 2\ sinx} = f\left( 0 \right)\]
\[ \Rightarrow \lim_{x \to 0} \frac{x\left( 2 + 3\frac{\ sinx}{x} \right)}{x\left( 3 + 2\frac{\ sinx}{x} \right)} = f\left( 0 \right)\]
\[ \Rightarrow \lim_{x \to 0} \frac{\left( 2 + 3\frac{\ sinx}{x} \right)}{\left( 3 + 2\frac{\ sinx}{x} \right)} = f\left( 0 \right)\]
\[ \Rightarrow \frac{\lim_{x \to 0} \left( 2 + 3\frac{\ sinx}{x} \right)}{\lim_{x \to 0} \left( 3 + 2\frac{\ sinx}{x} \right)} = f\left( 0 \right)\]
\[ \Rightarrow \frac{2 + 3 \lim_{x \to 0} \left( \frac{\ sinx}{x} \right)}{3 + 2 \lim_{x \to 0} \left( \frac{\ sinx}{x} \right)} = f\left( 0 \right)\]
\[ \Rightarrow \frac{2 + 3 \times 1}{3 + 2 \times 1} = f\left( 0 \right)\]
\[ \Rightarrow \frac{5}{5} = f\left( 0 \right)\]
\[ \Rightarrow f\left( 0 \right) = 1\]
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