Key Concepts and Formulas

• Definition of $f(x)$: The function $f(x)$ is the signum function, $\text{sgn}(x)$, which is $-1$ for $x<0$, $1$ for $x>0$, and $1$ for $x=0$.
• Definition of $g(x)$: The function $g(x)$ is defined as $\frac{\sin(x+1)}{x+1}$ for $x \neq -1$ and $1$ for $x=-1$. This resembles the limit definition of $\frac{\sin u}{u}$ as $u \to 0$. Specifically, $\lim_{u \to 0} \frac{\sin u}{u} = 1$.
• Limit of Composite Functions: To evaluate $\lim_{x \to a} g(F(x))$, we first evaluate the limit of the inner function, $L = \lim_{x \to a} F(x)$. Then, if $g$ is continuous at $L$, the limit is $g(L)$. If $g$ is not continuous at $L$, we need to analyze the behavior of $F(x)$ as it approaches $L$ from different sides.
• Greatest Integer Function $[x]$: The greatest integer function $[x]$ gives the largest integer less than or equal to $x$.

Step-by-Step Solution

Step 1: Analyze the function $f(x)$ The function $f(x)$ is defined as: $f(x)=\left\{\begin{array}{cc}\frac{x}{|x|}, & x \neq 0 \\ 1, & x=0\end{array}\right.$ For $x > 0$, $|x| = x$, so $f(x) = \frac{x}{x} = 1$. For $x < 0$, $|x| = -x$, so $f(x) = \frac{x}{-x} = -1$. Thus, $f(x)$ is the signum function: $f(x)=\left\{\begin{array}{cc}1, & x > 0 \\ -1, & x < 0 \\ 1, & x=0\end{array}\right.$ This can be written as $f(x) = \text{sgn}(x)$ for $x \neq 0$, and $f(0)=1$.

Step 2: Analyze the function $h(x)$ The function $h(x)$ is given by $h(x) = 2[x] - f(x)$. We need to evaluate $\lim\limits_{x \rightarrow 1} g(h(x-1))$. Let $y = x-1$. As $x \to 1$, $y \to 0$. We are interested in the behavior of $h(y)$ as $y \to 0$.

Let's analyze $h(x-1)$ as $x \to 1$. This means we need to analyze $h(y)$ as $y \to 0$. We need to consider the limits from the right ($x \to 1^+$) and from the left ($x \to 1^-$).

Step 3: Evaluate $h(x-1)$ as $x \to 1^+$ If $x \to 1^+$, then $x$ is slightly greater than 1. Let $x = 1 + \epsilon$, where $\epsilon$ is a small positive number. Then $x-1 = \epsilon$. So we are interested in $h(\epsilon)$ as $\epsilon \to 0^+$. For $x \to 1^+$, $[x] = [1+\epsilon] = 1$. And for $x \to 1^+$, $x > 0$, so $f(x) = 1$. Therefore, for $x \to 1^+$, $h(x-1) = h(\epsilon) = 2[\epsilon] - f(\epsilon)$. Since $\epsilon \to 0^+$, $[\epsilon] = 0$. Since $\epsilon > 0$, $f(\epsilon) = 1$. So, $h(x-1) = 2(0) - 1 = -1$ as $x \to 1^+$.

Step 4: Evaluate $h(x-1)$ as $x \to 1^-$ If $x \to 1^-$, then $x$ is slightly less than 1. Let $x = 1 - \epsilon$, where $\epsilon$ is a small positive number. Then $x-1 = -\epsilon$. So we are interested in $h(-\epsilon)$ as $\epsilon \to 0^+$. For $x \to 1^-$, $[x] = [1-\epsilon] = 0$. And for $x \to 1^-$, $x > 0$, so $f(x) = 1$. Therefore, for $x \to 1^-$, $h(x-1) = h(-\epsilon) = 2[-\epsilon] - f(-\epsilon)$. Since $\epsilon \to 0^+$, $-\epsilon$ is a small negative number. So, $[-\epsilon] = -1$. Since $-\epsilon < 0$, $f(-\epsilon) = -1$. So, $h(x-1) = 2(-1) - (-1) = -2 + 1 = -1$ as $x \to 1^-$.

Step 5: Evaluate the limit of $g(h(x-1))$ as $x \to 1^+$ From Step 3, as $x \to 1^+$, $h(x-1) \to -1$. We need to evaluate $\lim\limits_{x \rightarrow 1^+} g(h(x-1))$. Let $u = h(x-1)$. As $x \to 1^+$, $u \to -1$. The function $g(u)$ is defined as: $g(u)=\left\{\begin{array}{cc}\frac{\sin (u+1)}{(u+1)}, & u \neq-1 \\ 1, & u=-1\end{array}\right.$ We are considering the limit as $u \to -1$. We have $\lim\limits_{u \rightarrow -1} \frac{\sin (u+1)}{(u+1)}$. Let $v = u+1$. As $u \to -1$, $v \to 0$. So, $\lim\limits_{v \rightarrow 0} \frac{\sin v}{v} = 1$. Since $h(x-1)$ approaches $-1$ as $x \to 1^+$, and $g(u)$ is defined to be $1$ at $u=-1$, the limit is: $\lim\limits_{x \rightarrow 1^+} g(h(x-1)) = g(\lim\limits_{x \rightarrow 1^+} h(x-1)) = g(-1) = 1$. Alternatively, since $h(x-1)$ approaches $-1$ but is not necessarily equal to $-1$ in the limit process, we use the definition of $g(u)$ for $u \neq -1$: $\lim\limits_{x \rightarrow 1^+} g(h(x-1)) = \lim\limits_{u \rightarrow -1^+} \frac{\sin (u+1)}{(u+1)} = 1$. (Note: The original solution incorrectly stated $h(x-1) = 0-1 = -1$ for $x \to 1^+$ which implies $[x]=1$ and $f(x)=1$, leading to $h(x-1) = 2[x-1] - f(x-1)$. Let's re-evaluate $h(x-1)$ carefully).

Let's re-evaluate $h(x-1)$ more rigorously for the limit $x \to 1$. We are interested in $h(y)$ where $y = x-1$. As $x \to 1$, $y \to 0$. We need to evaluate $\lim_{y \to 0} h(y) = \lim_{y \to 0} (2[y] - f(y))$.

Case 1: $y \to 0^+$ (i.e., $x \to 1^+$) As $y \to 0^+$, $[y] = 0$. As $y \to 0^+$, $y > 0$, so $f(y) = 1$. So, $\lim_{y \to 0^+} h(y) = 2(0) - 1 = -1$.

Case 2: $y \to 0^-$ (i.e., $x \to 1^-$) As $y \to 0^-$, $[y] = -1$. As $y \to 0^-$, $y < 0$, so $f(y) = -1$. So, $\lim_{y \to 0^-} h(y) = 2(-1) - (-1) = -2 + 1 = -1$.

Since $\lim_{y \to 0^+} h(y) = -1$ and $\lim_{y \to 0^-} h(y) = -1$, we have $\lim_{y \to 0} h(y) = -1$. This means $\lim_{x \to 1} h(x-1) = -1$.

Step 6: Evaluate the limit of $g(h(x-1))$ We need to find $\lim\limits_{x \rightarrow 1} g(h(x-1))$. Let $u = h(x-1)$. As $x \to 1$, we found that $u \to -1$. We need to evaluate $\lim\limits_{u \rightarrow -1} g(u)$. The function $g(u)$ is defined as: $g(u)=\left\{\begin{array}{cc}\frac{\sin (u+1)}{(u+1)}, & u \neq-1 \\ 1, & u=-1\end{array}\right.$ We are interested in the limit as $u \to -1$. We know that $\lim\limits_{u \rightarrow -1} \frac{\sin (u+1)}{(u+1)}$. Let $v = u+1$. As $u \to -1$, $v \to 0$. The limit becomes $\lim\limits_{v \rightarrow 0} \frac{\sin v}{v} = 1$.

Since $h(x-1)$ approaches $-1$ as $x \to 1$, and the limit of $g(u)$ as $u \to -1$ is $1$, we have: $\lim\limits_{x \rightarrow 1} g(h(x-1)) = \lim\limits_{u \rightarrow -1} g(u) = 1$.

The original solution's calculation for $h(x-1)$ was slightly convoluted. Let's break down the $h(x-1)$ calculation as per the original solution's structure. If $x \to 1^+$, then $x=1+\epsilon$ for small $\epsilon > 0$. $h(x-1) = h(\epsilon) = 2[\epsilon] - f(\epsilon)$. Since $\epsilon \to 0^+$, $[\epsilon]=0$ and $f(\epsilon)=1$. So $h(\epsilon) = 2(0) - 1 = -1$. If $x \to 1^-$, then $x=1-\epsilon$ for small $\epsilon > 0$. $h(x-1) = h(-\epsilon) = 2[-\epsilon] - f(-\epsilon)$. Since $\epsilon \to 0^+$, $-\epsilon \to 0^-$. So $[-\epsilon]=-1$ and $f(-\epsilon)=-1$. So $h(-\epsilon) = 2(-1) - (-1) = -2 + 1 = -1$. Thus, in both cases, $h(x-1) \to -1$ as $x \to 1$.

Now, we need to evaluate $\lim\limits_{x \rightarrow 1} g(h(x-1))$. Let $u = h(x-1)$. As $x \to 1$, $u \to -1$. We are evaluating $\lim\limits_{u \rightarrow -1} g(u)$. $g(u) = \frac{\sin(u+1)}{u+1}$ for $u \neq -1$. So, $\lim\limits_{u \rightarrow -1} g(u) = \lim\limits_{u \rightarrow -1} \frac{\sin(u+1)}{u+1}$. Let $v = u+1$. As $u \to -1$, $v \to 0$. The limit becomes $\lim\limits_{v \rightarrow 0} \frac{\sin v}{v} = 1$.

Common Mistakes & Tips

• Careful analysis of the greatest integer function: Pay close attention to whether the argument of $[x]$ is approaching an integer from the left or right, as this determines the value of $[x]$. For example, $[0^+]=0$ but $[0^-]=-1$.
• Understanding function composition in limits: When dealing with $\lim_{x \to a} g(F(x))$, first find the limit of the inner function $L = \lim_{x \to a} F(x)$. Then, evaluate $g(L)$ if $g$ is continuous at $L$. If $g$ is not continuous at $L$, analyze the behavior of $F(x)$ as it approaches $L$.
• Signum function definition: Remember that $f(0)=1$, not 0, which is a common point of confusion.

Summary

To find the limit of the composite function $g(h(x-1))$ as $x \to 1$, we first analyzed the behavior of the inner function $h(x-1)$ as $x \to 1$. We found that as $x$ approaches 1 from both the left and the right, $h(x-1)$ approaches $-1$. Then, we evaluated the limit of the outer function $g(u)$ as $u$ approaches $-1$. Using the definition of $g(u)$ and the standard limit $\lim_{v \to 0} \frac{\sin v}{v} = 1$, we determined that the limit of $g(h(x-1))$ as $x \to 1$ is 1.

The final answer is $\boxed{1}$.