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JEE Mains · Maths · STD 12 - 1. relation and function

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

  1. A \(1\)
  2. B \(-1\)
  3. C \(-1\)
  4. D \(0\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(1\)

Step-by-step Solution

Detailed explanation

\(LHL =\lim _{ k \rightarrow 0} g ( h (- k )) , k > 0\) \(=\lim _{ k \rightarrow 0} g (-2+1) \because f ( x )=-1 \forall x < 0\) \(= g (-1)=1\) \(RHL =\lim _{ k \rightarrow 0} g ( h ( k )) , k > 0\) \(=\lim _{ k \rightarrow 0} g (-1) , \because f ( x )=1, \forall x > 0\) \(=1\)