Question

slove for me plz....​

Answer 1

Given :-

$\rm f(x) = \begin{cases} \rm x ,when \: 0 \leqslant x \leqslant 1\\ \\ \rm2 - x ,\: when \: 1 < x \leqslant 2& \end{cases}$

To find :-

$\displaystyle\lim_{\rm \: x \to \: 1}{ \rm \: f(x)}$

Solution :-

$\displaystyle\lim_{\rm \: x \to \: 1^{ - } }{ \rm \: f(x)} = \displaystyle\lim_{\rm \: x \to \: 1^{ + }}{ \rm \: f(x)} = \displaystyle\lim_{\rm \: x \to \: 1}{ \rm \: f(x)}$

$\longrightarrow\displaystyle\lim_{\rm \: x \to \: 1^{ - } }{ \rm \: f(x)} = \displaystyle\lim_{\rm x \to \: 1^{ - } } \rm x \\ \\ \longrightarrow\displaystyle\lim_{\rm \: x \to \: 1^{ - } }{ \rm \: x} = 1$

$\longrightarrow\displaystyle\lim_{\rm \: x \to \: 1^{ + } }{ \rm \: f(x)} = \displaystyle\lim_{\rm x \to \: 1^{ + } } \rm 2 - x \\ \\ \longrightarrow \displaystyle\lim_{\rm x \to \: 1^{ +} } \rm 2 - x = 2 - 1\\ \\ \longrightarrow 1$

$\therefore \: \displaystyle\lim_{\rm \: x \to \: 1^{ - } }{ \rm \: f(x)} = \displaystyle\lim_{\rm \: x \to \: 1^{ +}}{ \rm \: f(x)} = \displaystyle\lim_{\rm \: x \to \: 1}{ \rm \: f(x)} = 1$

Answer :-

Option ( a) 1 is correct

Answer 2

Step-by-step explanation:

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