Question

Guess the value of the limit (if it exists) by evaluating the function at the given numbers (correct to six decimal places).

\(\begin{array}{l}\mathop {lim}\limits_{x \to 0} \frac{{sinx}}{{x + tanx}}\\x = \pm 1, \pm 0.5, \pm 0.2, \pm 0.1, \pm 0.05, \pm 0.01\end{array}\)

Short answer

The value of \(\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{{x + \tan x}}\) is \(\frac{1}{2}\).

Step-by-step solution

Introduction

The function does not exist on the given limit but it can sustain on the nearby values of the limit. To find the value of the limit we have to find the nearby limit values.

Given information

The given function and the values are,

\(\begin{array}{l}\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{{x + \tan x}}\\x = \pm 1, \pm 0.5, \pm 0.2, \pm 0.1, \pm 0.05, \pm 0.01\end{array}\)

Explanation

\(\)

The following table is for \(\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{{x + \tan x}}\)with the values of \(x = \pm 1, \pm 0.5, \pm 0.2, \pm 0.1, \pm 0.05, \pm 0.01\).

\(\)

x	f(x)
\(\begin{array}{l} \pm 1\\ \pm 0.5\\ \pm 0.2\\ \pm 0.1\\ \pm 0.05\\ \pm 0.01\end{array}\) \(\)	\(\begin{array}{l}0.329033\\0.458209\\0.493331\\0.498333\\0.499583\\0.499983\end{array}\)

It seems the value of the limit

\(\begin{array}{c}\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{{x + \tan x}} = 0.5\\ = \frac{1}{2}\end{array}\)

Hence, the limit value of the given function is, \(\frac{1}{2}\).