Question

Solve the given expression, the given expression is:
\[\mathop {\lim }\limits_{x \to 0} \left[ {\dfrac{x}{{\sin x}}} \right] + {x^x}\]where [.] is the greatest integer function?
A. 1
B. 2
C. 0
D. -1

Answer 1

Hint:The given question is to solve for the limit function, here we have to use the properties of limit in order to get the solution. And here the greatest integer function is also used according to which the value of the decimal number is the first integer number behind it on the number line. For example, the value of 0.6 in the greatest integer function will be zero.

Formulae Used:
\[\mathop {\lim }\limits_{x \to 0} \left[ {\dfrac{{\sin x}}{x}} \right] = 1\]

Complete step-by-step solution:
Here to solve the given question we need to use the properties of limit, in which we know that when putting the value of limit the term is defined then directly we can have the answer for the limit function, and the terms which give in any indeterminate values then we have to use some more properties then, here we have:
\[
   \Rightarrow \mathop {\lim }\limits_{x \to 0} \left[ {\dfrac{x}{{\sin x}}} \right] + \mathop {\lim }\limits_{x \to 0} {x^x} \\
   \Rightarrow \mathop {\lim }\limits_{x \to 0} \left[ {\dfrac{1}{{\dfrac{{\sin x}}{x}}}} \right] + \mathop {\lim }\limits_{x \to 0} \left( {{0^0}} \right) \\
   \Rightarrow \left[ {\dfrac{1}{{\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x}}}} \right] + 1 = \left[ {\dfrac{1}{1}} \right] + 1 = 2\,(\because [1] = 1) \\
 \]
Here we got the value of the expression as two.

Note:Here in the given question we use the properties of the limit in order to solve for the value of the function, here we already knew that \[\mathop {\lim }\limits_{x \to 0} \left[ {\dfrac{{\sin x}}{x}} \right] = 1\]reciprocating it and bringing it in the form of \[\mathop {\lim }\limits_{x \to 0} \left[ {\dfrac{x}{{\sin x}}} \right] = 1\], also we must know that the outcome of any number power to 0 is equal to 1, hence addition of both of these is 2.