Question

$\lim_{x \to 0} \frac{|x|}{|x| + x^2} =$}

• A. 0
• B. 1
• C. -1
• D. $\frac{1}{2}$

Hint:

Since $x^2 = |x|^2$, you can rewrite the expression as $|x|/(|x| + |x|^2)$ and cancel $|x|$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
To evaluate a limit involving absolute values at a point where the argument is zero, we must evaluate the left-hand limit (LHL) and the right-hand limit (RHL) separately.
Step 2: Key Formula or Approach:
RHL: Let $x \to 0^+$, which means $x>0$. Therefore, $|x| = x$.
LHL: Let $x \to 0^-$, which means $x<0$. Therefore, $|x| = -x$.
If RHL = LHL, the limit exists and equals that value.
Step 3: Detailed Explanation:
Evaluate the Right-Hand Limit (RHL):
As $x \to 0^+$, $|x| = x$.
\[ \text{RHL} = \lim_{x \to 0^+} \frac{x}{x + x^2} \] Divide numerator and denominator by $x$ (valid since $x \neq 0$ inside the limit):
\[ \text{RHL} = \lim_{x \to 0^+} \frac{1}{1 + x} = \frac{1}{1 + 0} = 1 \] Evaluate the Left-Hand Limit (LHL):
As $x \to 0^-$, $|x| = -x$.
\[ \text{LHL} = \lim_{x \to 0^-} \frac{-x}{-x + x^2} \] Divide numerator and denominator by $x$:
\[ \text{LHL} = \lim_{x \to 0^-} \frac{-1}{-1 + x} = \frac{-1}{-1 + 0} = \frac{-1}{-1} = 1 \] Since $\text{RHL} = \text{LHL} = 1$, the overall limit exists and is equal to 1.
Step 4: Final Answer:
The limit is 1.