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Prove that the function f ( x ) = { sin x x , x < 0 x + 1 , x ≥ 0 is everywhere continuous.

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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.

 

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उत्तर

When x < 0, we have  

\[f\left( x \right) = \frac{\text{ sin } x}{x}\]

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

\[f\left( x \right) = x + 1\]

Therefore,

\[f\left( x \right)\] is continuous at each x > 0.

Given:

\[f\left( x \right) = \binom{\frac{\text{ sin } x}{x}, x < 0}{x + 1, x \geq 0}\]
We have
(LHL at x = 0) = \[\lim_{x \to 0^-} f\left( x \right) = \lim_{h \to 0} f\left( 0 - h \right) = \lim_{h \to 0} f\left( - h \right) = \lim_{h \to 0} \left( \frac{\sin\left( - h \right)}{- h} \right) = \lim_{h \to 0} \left( \frac{\sin\left( h \right)}{h} \right) = 1\]
(RHL at x = 0) =
\[\lim_{x \to 0^+} f\left( x \right) = \lim_{h \to 0} f\left( 0 + h \right) = \lim_{h \to 0} f\left( h \right) = \lim_{h \to 0} \left( h + 1 \right) = 1\]
Also, 
\[f\left( 0 \right) = 0 + 1 = 1\]
\[\lim_{x \to 0^-} f\left( x \right) = \lim_{x \to 0^+} f\left( x \right) = f\left( 0 \right)\]
Thus
\[f\left( x \right)\]  is continuous at x = 0.
Hence,
\[f\left( x \right)\] is everywhere continuous.
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अध्याय 8: Continuity - Exercise 9.2 [पृष्ठ ३४]

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आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 8 Continuity
Exercise 9.2 | Q 1 | पृष्ठ ३४

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