Question

$f(x)=\left\\{\begin{array}{ll}\frac{x}{2 x^{2}+|x|} & , x \neq 0 \\ 1 & , x=0\end{array}\right.\( then \)f(x)$ is (A) Continuous but non-differentiable at \(\mathrm{x}=0\) (B) Differentiable at \(\mathrm{x}=0\) (C) Discontinuous at \(\mathrm{x}=0\) (D) None of these

Short answer

Question: Determine the properties of the function \(f(x) = \frac{x}{2x^{2} \pm |x|}\) at \(x=0\). A. Continuous and differentiable at \(x=0\) B. Continuous and not differentiable at \(x=0\) C. Discontinuous at \(x=0\) D. None of the above Answer: C. Discontinuous at \(x=0\).

Step-by-step solution

Check Continuity at \(x=0\)

To check the continuity of \(f(x)\) at \(x=0\), we need to verify if the following condition holds: $$\lim_{x \to 0^{-}} f(x) = f(0) = \lim_{x \to 0^{+}} f(x)$$ Given that \(f(0) = 1\), let's find the left and right limits.

Left Limit (\(x \to 0^{-}\)) Calculation

As \(x \to 0^{-}\), we know that \(x\) is negative, and therefore \(|x| = -x\). Then, we can rewrite the given function as: $$f(x) = \frac{x}{2x^{2} - x} \quad \text{for} \quad x < 0$$ Now, we can find the left limit of \(f(x)\) as \(x \to 0^{-}\): $$\lim_{x \to 0^{-}} \frac{x}{2x^{2} - x}$$ Since both the numerator and denominator go to \(0\) as \(x \to 0^{-}\), we can use L'Hopital's rule: $$\lim_{x \to 0^{-}} \frac{\frac{d}{dx} x}{\frac{d}{dx}(2x^{2} - x)} = \lim_{x \to 0^{-}} \frac{1}{4x - 1}$$ As \(x \to 0^{-}\), the limit becomes: $$\lim_{x \to 0^{-}} \frac{1}{4x - 1} = \frac{1}{-1} = -1$$

Right Limit (\(x \to 0^{+}\)) Calculation

Since \(x \to 0^{+}\), \(x\) is positive, and therefore, \(|x| = x\). We can rewrite the given function as: $$f(x) = \frac{x}{2x^{2} + x} \quad \text{for} \quad x > 0$$ Now, we can find the right limit of \(f(x)\) as \(x \to 0^{+}\): $$\lim_{x \to 0^{+}} \frac{x}{2x^{2} + x}$$ Again, we use L'Hopital's rule to evaluate the limit: $$\lim_{x \to 0^{+}} \frac{\frac{d}{dx} x}{\frac{d}{dx}(2x^{2} + x)} = \lim_{x \to 0^{+}} \frac{1}{4x + 1}$$ As \(x \to 0^{+}\), the limit becomes: $$\lim_{x \to 0^{+}} \frac{1}{4x + 1} = \frac{1}{1} = 1$$

Comparing Left and Right Limits, and Function Value at \(x=0\)

We have found the left limit, the right limit, and the function's value at \(x=0\) as follows: $$\lim_{x \to 0^{-}}f(x) = -1$$ $$\lim_{x \to 0^{+}}f(x) = 1$$ $$f(0) = 1$$ Since the left and right limits of the function do not match, the function \(f(x)\) is not continuous at \(x=0\). Therefore, we can conclude that the function is discontinuous at \(x=0\). The correct answer is option (C) Discontinuous at \(\mathrm{x}=0\).