Let $f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {\left( {\frac{1}{2}} \right)\left\{ {g\left( x \right) + \left( x \right)} \right\}\sin \left( x \right),x \ge 1}\\ {\sin x/x,\,\,\,x 0}\\ { - 1,{\rm{if}}\,\,x < 0}\\ {0,{\rm{if}}\,\,x = 0} \end{array}} \right.$ Then, $\mathop {\lim }\limits_{x \to 1} f\left( x \right)$ is equal to

JEE MATHEMATICS MCQ: LIMITS

1 Easy

1 of 1

Analyzing +4 | -1

Let

f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {\left( {\frac{1}{2}} \right)\left\{ {g\left( x \right) + \left( x \right)} \right\}\sin \left( x \right),x \ge 1}\\ {\sin x/x,\,\,\,x < 1} \end{array}} \right.

Where

g\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {1,\mathrm{if}\,\,x > 0}\\ { - 1,\mathrm{if}\,\,x < 0}\\ {0,\mathrm{if}\,\,x = 0} \end{array}} \right.

Then,

\lim\limits_{x \to 1} f\left( x \right)

is equal to

View answer and explanation

Correct option: C

\sin 1

Option C (sin 1) is correct because both the left-hand limit and right-hand limit of the function as x approaches 1 are equal to sin 1.