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Question
If \[f\left( x \right) = \left\{ \begin{array}{l}2x + 3, & x \leq 0 \\ 3 \left( x + 1 \right), & x > 0\end{array} . \right.\] find \[\lim_{x \to 1} f\left( x \right)\]
Solution
\[f\left( x \right) = \left\{ \begin{array}{l}2x + 3, & x \leq 0 \\ 3 \left( x + 1 \right), & x > 0\end{array} . \right.\] find \[\lim_{x \to 1} f\left( x \right)\]
∴ `lim_(x->0^-) f(x) = lim_(x->0^-) (2x + 3) = 2 xx 0 + 3 = 3`
and `lim_(x->0^+) f(x) = lim_(x->0^+) 3 (x + 1)= 3(0 + 1)=3`
So, `lim_(x->0) f(x)` exists and is equal to 3.
`lim_(x->1^-) f(x) = lim_(x->1^-) 2x + 3 = 2 xx 1 + 3 = 5`
`lim_(x->1^+) f(x) = lim_(x->1^+) 3(x + 1)= 3(1 + 1)=6`
`lim_(x->1^-) f(x) ne lim_(x->1^+) f(x)`
Hence, `lim_(x->1) f(x)` does not exist
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