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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

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

Mathematics
NCERT
Standard XI

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