Key Concepts and Formulas

• Greatest Integer Function [x]: The greatest integer less than or equal to x. For x approaching 0 from the left ($x \to 0^-$), x is a small negative number. Thus, [x] will be -1.
• Absolute Value Function |x|: For x approaching 0 from the left ($x \to 0^-$), x is a small negative number. Thus, |x| = -x.
• Limits of Trigonometric Functions: The limit of $\sin(y)$ as $y \to c$ is $\sin(c)$.
• Limit Evaluation: When evaluating a limit of a function $f(x)$ as $x \to a$, we substitute values of x that are arbitrarily close to 'a' but not equal to 'a'. For one-sided limits, we consider values from only one side.

Step-by-Step Solution

Step 1: Analyze the limit expression and the behavior of functions as $x \to 0^-$. We are asked to evaluate the limit: $L = \mathop {\lim }\limits_{x \to {0^ - }} \,\,{{x\left( {\left[ x \right] + \left| x \right|} \right)\sin \left[ x \right]} \over {\left| x \right|}}$ As $x \to 0^-$, x is a small negative number. Therefore, we can analyze the terms:

• $[x]$: Since x is slightly less than 0, the greatest integer less than or equal to x is -1. So, $[x] = -1$.
• $|x|$: Since x is negative, $|x| = -x$.
• $\sin[x]$: Since $[x] = -1$, $\sin[x] = \sin(-1)$. Using the property $\sin(-\theta) = -\sin(\theta)$, we get $\sin(-1) = -\sin(1)$.

Step 2: Substitute the determined values of $[x]$ and $|x|$ into the limit expression. Replacing $[x]$ with -1 and $|x|$ with -x in the expression: $L = \mathop {\lim }\limits_{x \to {0^ - }} \,\,{{x\left( {(-1) + (-x)}\right)\sin \left( -1 \right)} \over {\left( -x \right)}}$ Simplify the expression inside the parenthesis: $L = \mathop {\lim }\limits_{x \to {0^ - }} \,\,{{x\left( {-1 - x}\right)\sin \left( -1 \right)} \over {\left( -x \right)}}$

Step 3: Simplify the expression by canceling out common terms. We can cancel out 'x' from the numerator and the denominator, provided $x \neq 0$, which is true as we are considering a limit as $x \to 0^-$. $L = \mathop {\lim }\limits_{x \to {0^ - }} \,\,{{(-1 - x)\sin \left( -1 \right)} \over {\left( -1 \right)}}$ Now, we can simplify the denominator: $L = \mathop {\lim }\limits_{x \to {0^ - }} \,\,{{(-1 - x)\sin \left( -1 \right)} \over {-1}}$ Multiply the numerator by -1 (from the denominator): $L = \mathop {\lim }\limits_{x \to {0^ - }} \,\,{(1 + x)\sin \left( -1 \right)}$

Step 4: Evaluate the limit by substituting the value of x. As $x \to 0^-$, the term 'x' approaches 0. We can directly substitute x = 0 into the simplified expression: $L = (1 + 0)\sin \left( -1 \right)$ $L = (1)\sin \left( -1 \right)$ $L = \sin \left( -1 \right)$ Using the property $\sin(-\theta) = -\sin(\theta)$: $L = -\sin(1)$

Common Mistakes & Tips

• Incorrectly evaluating [x] for $x \to 0^-$: Many students might mistake $[x]$ for 0 when $x \to 0^-$. Remember that for any negative number slightly greater than -1 (e.g., -0.001), the greatest integer less than or equal to it is -1.
• Forgetting the negative sign in $\sin(-1)$: The sine function is odd, meaning $\sin(-\theta) = -\sin(\theta)$. Failing to account for this will lead to an incorrect sign in the final answer.
• Algebraic errors when simplifying: Carefully simplify the expression by canceling terms and handling the negative signs correctly. It's often helpful to write out the steps explicitly.

Summary

To evaluate the given limit as $x$ approaches 0 from the left, we first analyzed the behavior of the greatest integer function $[x]$ and the absolute value function $|x|$ for small negative values of $x$. We found that as $x \to 0^-$, $[x] = -1$ and $|x| = -x$. Substituting these into the expression, we simplified the fraction by canceling out the $x$ terms. Finally, we evaluated the limit by substituting $x=0$ into the simplified expression and using the property of the sine function for negative angles, arriving at $-\sin(1)$.

The final answer is $\boxed{-\sin 1}$.