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If $f: \mathbf{R} \rightarrow \mathbf{R}$ and $g: \mathbf{R} \rightarrow \mathbf{R}$ be defined by
$\begin{aligned}
& f(x)=\left\{\begin{array}{cc}
x+2, & x>0 \\
2-x, & x \leq 0
\end{array}\right. \text { and } \\
& g(x)=\left\{\begin{array}{cc}
x^2-2 x-2, & 1 \leq x < 2 \\
x-7 & x \geq 2 \\
x+5, & x < 1
\end{array} \text { then } \lim _{x \rightarrow 0} g \circ f(x)\right.
\end{aligned}$
Options:
$\begin{aligned}
& f(x)=\left\{\begin{array}{cc}
x+2, & x>0 \\
2-x, & x \leq 0
\end{array}\right. \text { and } \\
& g(x)=\left\{\begin{array}{cc}
x^2-2 x-2, & 1 \leq x < 2 \\
x-7 & x \geq 2 \\
x+5, & x < 1
\end{array} \text { then } \lim _{x \rightarrow 0} g \circ f(x)\right.
\end{aligned}$
Solution:
2563 Upvotes
Verified Answer
The correct answer is:
is equal to -5
$\begin{aligned} & f(x)=\left\{\begin{array}{cc}x+2, & x>0 \\ 2-x, & x \leq 0\end{array}\right. \\ & g(x)=\left\{\begin{array}{cc}x^2-2 x-2, & 1 \leq x < 2 \\ x-7, & x \geq 2 \\ x+5, & x < 1\end{array}\right.\end{aligned}$
$\begin{gathered}g(f(x))=\left\{\begin{array}{cc}(f(x))^2-2 f(x)-2, & 1 \leq f(x) < 2 \\ f(x)-7, & f(x) \geq 2 \\ f(x)+5, & f(x) < 1\end{array}\right. \\ f(0)=2, \lim _{x \rightarrow 0} g f(x)=g f(0)=g(2)=2-7=-5\end{gathered}$
$\begin{gathered}g(f(x))=\left\{\begin{array}{cc}(f(x))^2-2 f(x)-2, & 1 \leq f(x) < 2 \\ f(x)-7, & f(x) \geq 2 \\ f(x)+5, & f(x) < 1\end{array}\right. \\ f(0)=2, \lim _{x \rightarrow 0} g f(x)=g f(0)=g(2)=2-7=-5\end{gathered}$
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