Question

Let $f, g$ and $h$ be the real valued functions defined on $R$ as $f(x)=\begin{cases} \frac{x}{|x|}, & x \neq 0 \\1, & x=0\end{cases}, g(x)=\begin{cases} \frac{\sin (x+1)}{(x+1)}, & x \neq-1 \\1, & x=-1\end{cases}$ and $h(x)=2[x]-f(x)$, where $[x]$ is the greatest integer $\leq x$ Then the value of $\displaystyle\lim _{x \rightarrow 1} g(h(x-1))$ is

• A. 1
• B. 0
• C. $\sin (1)$
• D. $-1$

Answer 1

The Correct Option is A

To find the value of \(\lim_{x \rightarrow 1} g(h(x-1))\), let's first understand the functions involved:

1. The function \(f(x)\) is defined as:
   • For \(x \neq 0\), \(f(x) = \frac{x}{|x|}\). This is the sign function, which gives \(1\) for positive \(x\) and \(-1\) for negative \(x\).
   • For \(x = 0\), \(f(x) = 1\).
2. The function \(g(x)\) is defined as:
   • For \(x \neq -1\), \(g(x) = \frac{\sin (x+1)}{x+1}\), which is continuous and equals \(1\) when \(x = -1\) (using L'Hôpital's Rule).
   • For \(x = -1\), \(g(x) = 1\).
3. The function \(h(x)\) is defined as \(h(x) = 2[x] - f(x)\), where \([x]\) is the greatest integer ≤ \(x\).

Now, evaluate \(h(x-1)\) as \(x \rightarrow 1\):

• As \(x \rightarrow 1\), \(x-1 \rightarrow 0\). Therefore, \([x-1] = \lfloor 0 \rfloor = 0\).
• Since \(x-1\) approaches 0 from the positive side, \(f(x-1) = f(0^{+}) = 1\).
• So, \(h(x-1) = 2 \times [x-1] - f(x-1) = 2 \times 0 - 1 = -1\).

Substitute \(h(x-1) = -1\) into \(g(x)\):

• Since \(h(x-1) = -1\), \(g(h(x-1)) = g(-1) = 1\).

Thus, the limit is:

\(\lim_{x \rightarrow 1} g(h(x-1)) = 1\).

The correct answer is 1.