JEE MATHEMATICS MCQ: LIMITS

1 Easy

1 of 1

Analyzing +4 | -1

Let

f, g

and

h

be the real valued functions defined of

\mathbb{R}

f(x)\left\{\begin{array}{ll} \frac{x}{|x|}, & x \neq 0 \\ 1, & x=0 \end{array}, \,\,\, g(x) \begin{cases}\frac{\sin (x+1)}{(x+1)}, & x \neq-1 \\ 1, & x=-1\end{cases}\right.

and

h(x)=2[x]-f(x)

, where

[x]

is the greatest integer

\leq x

.

Then the value of

\lim \limits_{{x \rightarrow 1}} g(h(x-1))

View answer and explanation

Correct option: A

Option A is correct because the limit of

g(h(x-1))

x

approaches 1 is 1.

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