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Question Number 27722 by NECx last updated on 13/Jan/18

find the limit of    f(x)= { ((1+x       x<1)),((k              x=0   c=0)),((1+x     , x>0)) :}

$${find}\:{the}\:{limit}\:{of} \\ $$ $$ \\ $$ $${f}\left({x}\right)=\begin{cases}{\mathrm{1}+{x}\:\:\:\:\:\:\:{x}<\mathrm{1}}\\{{k}\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}=\mathrm{0}\:\:\:{c}=\mathrm{0}}\\{\mathrm{1}+{x}\:\:\:\:\:,\:{x}>\mathrm{0}}\end{cases} \\ $$

Answered by NECx last updated on 15/Jan/18

  the limit is said to exist is the  LHL=RHL    LHL=lim_(x→1^− )  1+x=1+1=2  RHL=lim_(x→0^+ )  1+x=1+0=1    ∴Since the LHL≠RHL,then  the limit does not exist

$$\:\:{the}\:{limit}\:{is}\:{said}\:{to}\:{exist}\:{is}\:{the} \\ $$ $${LHL}={RHL} \\ $$ $$ \\ $$ $${LHL}=\underset{{x}\rightarrow\mathrm{1}^{−} } {\mathrm{lim}}\:\mathrm{1}+{x}=\mathrm{1}+\mathrm{1}=\mathrm{2} \\ $$ $${RHL}=\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\mathrm{1}+{x}=\mathrm{1}+\mathrm{0}=\mathrm{1} \\ $$ $$ \\ $$ $$\therefore{Since}\:{the}\:{LHL}\neq{RHL},{then} \\ $$ $${the}\:{limit}\:{does}\:{not}\:{exist} \\ $$ $$ \\ $$

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