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

Updated On: Mar 27, 2025
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  • $\sin (1)$
  • $-1$
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The Correct Option is A

Solution and Explanation

The correct answer is (A) : 1







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Concepts Used:

Application of Derivatives

Various Applications of Derivatives-

Rate of Change of Quantities:

If some other quantity ‘y’ causes some change in a quantity of surely ‘x’, in view of the fact that an equation of the form y = f(x) gets consistently pleased, i.e, ‘y’ is a function of ‘x’ then the rate of change of ‘y’ related to ‘x’ is to be given by 

\(\frac{\triangle y}{\triangle x}=\frac{y_2-y_1}{x_2-x_1}\)

This is also known to be as the Average Rate of Change.

Increasing and Decreasing Function:

Consider y = f(x) be a differentiable function (whose derivative exists at all points in the domain) in an interval x = (a,b).

  • If for any two points x1 and x2 in the interval x such a manner that x1 < x2, there holds an inequality f(x1) ≤ f(x2); then the function f(x) is known as increasing in this interval.
  • Likewise, if for any two points x1 and x2 in the interval x such a manner that x1 < x2, there holds an inequality f(x1) ≥ f(x2); then the function f(x) is known as decreasing in this interval.
  • The functions are commonly known as strictly increasing or decreasing functions, given the inequalities are strict: f(x1) < f(x2) for strictly increasing and f(x1) > f(x2) for strictly decreasing.

Read More: Application of Derivatives