Question

# Find $$\displaystyle \lim_{x\rightarrow 0}f\left ( x \right )$$ and $$\displaystyle \lim_{x\rightarrow 1}f\left ( x \right )$$ where $$f(x) =$$ $$\displaystyle \begin{cases}2x+3, \,\,\,\,\,\,\,x\leq 0 \\3\left ( x+1 \right ), \,\,\,\,x>0 \end{cases}$$

Solution

## $$f(x) = \displaystyle \begin{cases}2x+3, & \text{ } x\leq 0 \\3\left ( x+1 \right ), & \text{ } x>0 \end{cases}$$$$\displaystyle \lim_{x\rightarrow 0^{-}}f\left ( x \right )=\lim_{x\rightarrow 0}\left [ 2x+3 \right ]=2\left ( 0 \right )+3=3$$$$\displaystyle \lim_{x\rightarrow 0^{+}}f\left ( x \right )=\lim_{x\rightarrow 0}3\left ( x+1 \right )=3\left ( 0+1 \right )=3$$$$\displaystyle \therefore \lim_{x\rightarrow 0^{-}}f\left ( x \right )=\lim_{x\rightarrow 0}f\left ( x \right )=\lim_{x\rightarrow 0}f\left ( x =3\right )$$$$\displaystyle \lim_{x\rightarrow 1}f\left ( x \right )=\lim_{x\rightarrow 1}3\left ( x+3 \right )=3\left ( 1+1 \right )=6$$$$\displaystyle \lim_{x\rightarrow 1^{-}}f\left ( x \right )=\lim_{x\rightarrow 1}3\left ( x+1 \right )=3\left ( 1+1 \right )=6$$$$\displaystyle \lim_{x\rightarrow 1^{+}}f\left ( x \right )=\lim_{x\rightarrow 1}3\left ( x+1 \right )=3\left ( 1+1 \right )=6$$$$\displaystyle \therefore \lim_{x\rightarrow 1^{-}}f\left ( x \right )=\lim_{x\rightarrow 1^{+}}f\left ( x \right )=\lim_{x\rightarrow 1}f\left ( x \right )=6$$MathematicsNCERTStandard XI

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