Question

True or False \(\lim _ { x \rightarrow 0 } \frac { x + \sin x } { x } = 2 .\) Justify your answer.

Short answer

True, \(\lim _ { x \rightarrow 0 } \frac { x + \sin x } { x } = 2\).

Step-by-step solution

Recognize and simplify

Recognize that the limit is in the form where direct substitution would result in 0/0, an indeterminate form. Therefore, apply the property that \(\lim _ { x \rightarrow a } ( f(x) + g(x) ) = \lim _ { x \rightarrow a } f(x) + \lim _ { x \rightarrow a } g(x)\) to break it down as \(\lim _ { x \rightarrow 0 } \frac { x } { x } + \lim _ { x \rightarrow 0 } \frac { \sin x } { x }\)

Evaluate the limit

As x approaches 0, the first part \(\lim _ { x \rightarrow 0 } \frac { x } { x }\) simplifies to 1 since any number divided by itself equals 1 (with exception of dividing by zero). The second part \(\lim _ { x \rightarrow 0 } \frac { \sin x } { x }\) is a standard limit that also equals 1. This is a well-known limit that comes from the definition of the derivative of sine. Therefore, we get 1 + 1 = 2.

Interpret the result

The result tells us that, as x approaches 0, the quotient of the function \(x + \sin x\) and x approaches the limit 2, so the statement \(\lim _ { x \rightarrow 0 } \frac { x + \sin x } { x } = 2\) is true.