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प्रश्न
Prove that the function \[f\left( x \right) = \begin{cases}\frac{\sin x}{x}, & x < 0 \\ x + 1, & x \geq 0\end{cases}\] is everywhere continuous.
उत्तर
When x < 0, we have
We know that sin x as well as the identity function x are everywhere continuous. So, the quotient function
\[\frac{\text{ sin } x}{x}\]
When x > 0, we have
Therefore,
Given:
(LHL at x = 0) = \[\lim_{x \to 0^-} f\left( x \right) = \lim_{h \to 0} f\left( 0 - h \right) = \lim_{h \to 0} f\left( - h \right) = \lim_{h \to 0} \left( \frac{\sin\left( - h \right)}{- h} \right) = \lim_{h \to 0} \left( \frac{\sin\left( h \right)}{h} \right) = 1\]
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