MCQ
If $f(x) = \left\{ \begin{array}{l}\,\,\,\,\,\,\,x,\;{\rm{when\,\, }}0 \le x \le 1\\2 - x,\;{\rm{when \,\,}}1 < x \le 2\end{array} \right.$, then $\mathop {\lim }\limits_{x \to 1} f(x) = $
- ✓$1$
- B$2$
- C$0$
- DDoes not exist
✓
Answer
Correct option: A.
$1$
a
(a) Hence $\mathop {\lim }\limits_{x \to 1} \,f(x) = 1$
(a) Hence $\mathop {\lim }\limits_{x \to 1} \,f(x) = 1$
Aliter : $\mathop {\lim }\limits_{x \to 1 - } \,f(x) = \mathop {\lim }\limits_{h \to 0} \,\,(1 - h) = 1$
and $\mathop {\lim }\limits_{x \to 1 + } \,f(x) = \mathop {\lim }\limits_{h \to 0} \,\,2 - (1 + h) = 1$
Hence limit of function is $1$.
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