Let f, g and h be the real-valued functions d | vedclass

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(A) We need to find $\lim_{x \rightarrow 1} g(h(x-1))$. Let $t = x-1$. As $x \rightarrow 1$,$t \rightarrow 0$. So,we evaluate $\lim_{t \rightarrow 0}

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$1$

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(A) We need to find $\lim_{x \rightarrow 1} g(h(x-1))$. Let $t = x-1$. As $x \rightarrow 1$,$t \rightarrow 0$. So,we evaluate $\lim_{t \rightarrow 0} g(h(t))$.$h(t) = 2[t] - f(t)$.For $t \rightarrow 0^-$,$[t] = -1$ and $f(t) = \frac{t}{|t|} = -1$. Thus,$h(t) = 2(-1) - (-1) = -2 + 1 = -1$.Then,$\lim_{t \rightarrow 0^-} g(h(t)) = g(-1) = 1$.For $t \rightarrow 0^+$,$[t] = 0$ and $f(t) = \frac{t}{|t|} = 1$. Thus,$h(t) = 2(0) - 1 = -1$.Then,$\lim_{t \rightarrow 0^+} g(h(t)) = g(-1) = 1$.Since the left-hand limit and right-hand limit are equal,the limit is $1$.

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