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  • Let <img alt="\mathrm{f(x)= \begin{cases}x^2 & , x<1 \\ x & , 1<x<4, \text { then } \\ 4-x, & x>4\end{cases}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm

Let \mathrm{f(x)= \begin{cases}x^2 & , x<1 \\ x & , 1<x<4, \text { then } \\ 4-x, & x>4\end{cases}}
 

Option: 1

\mathrm{\lim _{x \rightarrow 1-} f(x)=10}
 


Option: 2

\mathrm{\lim _{x \rightarrow 1+} f(x)=12}
 


Option: 3

\mathrm{\lim _{x \rightarrow 4-} f(x)=4}
 


Option: 4

\mathrm{\lim _{x \rightarrow 4^{+}} f(x)=4}


Answers (1)

best_answer

Option (1): \mathrm{\lim _{x \rightarrow 1^{-}} f(x) =\lim _{h \rightarrow 0} f(1-h) }

                                 \mathrm{ =\lim _{h \rightarrow 0}(1-h)^2=1}

Option(2): \mathrm{ \lim _{x \rightarrow 1^{+}} f(x)= \lim _{h \rightarrow 0} f(1+h) }

                                \mathrm{=\lim _{h \rightarrow C}(1+h)=1}

Option (3): \mathrm{\lim _{x \rightarrow 4^{-}} f(x)= \lim _{h \rightarrow 0} f(4-h) }

                                 \mathrm{ =\lim _{h \rightarrow 0}(4-h)=4}

Option (4): \mathrm{\lim _{x \rightarrow 4^{+}} f(x)= \lim _{h \rightarrow 0} f(4+h) }

                                 \mathrm{ =\lim _{h \rightarrow 0} 4-(4+h)=0 }
Hence option 3 is correct.

 

Posted by

Rishi

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