Differentiate : y=(2x^2+1nsqrt (x)) lim _(xarrow 1)(x^3-1)/(x^2)-1 lim _(xarrow 3)((vert x-3vert )/(x-3))

Let's go through each limit step by step.<br /><br />### 1. Differentiate \( y = 2x^2 + \ln(\sqrt{x}) \)<br /><br />First, rewrite the function for easier differentiation:<br />\[ y = 2x^2 + \ln(x^{1/2}) \]<br />\[ y = 2x^2 + \frac{1}{2} \ln(x) \]<br /><br />Now, differentiate \( y \) with respect to \( x \):<br />\[ \frac{dy}{dx} = \frac{d}{dx} \left( 2x^2 + \frac{1}{2} \ln(x) \right) \]<br />\[ \frac{dy}{dx} = 4x + \frac{1}{2} \cdot \frac{1}{x} \]<br />\[ \frac{dy}{dx} = 4x + \frac{1}{2x} \]<br /><br />### 2. Evaluate \(\lim_{x \to 1} \frac{x^3 - 1}{x^2 - 1}\)<br /><br />Factor the numerator and denominator:<br />\[ x^3 - 1 = (x - 1)(x^2 + x + 1) \]<br />\[ x^2 - 1 = (x - 1)(x + 1) \]<br /><br />Cancel the common factor \((x - 1)\):<br />\[ \frac{x^3 - 1}{x^2 - 1} = \frac{(x - 1)(x^2 + x + 1)}{(x - 1)(x + 1)} = \frac{x^2 + x + 1}{x + 1} \]<br /><br />Now, evaluate the limit as \( x \to 1 \):<br />\[ \lim_{x \to 1} \frac{x^2 + x + 1}{x + 1} = \frac{1^2 + 1 + 1}{1 + 1} = \frac{3}{2} \]<br /><br />### 3. Evaluate \(\lim_{x \to 3} \frac{|x - 3|}{x - 3}\)<br /><br />Consider the definition of the absolute value:<br />- For \( x > 3 \), \( |x - 3| = x - 3 \)<br />- For \( x < 3 \), \( |x - 3| = -(x - 3) = 3 - x \)<br /><br />Since we are approaching \( x = 3 \) from the right (i.e., \( x \to 3^+ \)), \( |x - 3| = x - 3 \):<br />\[ \lim_{x \to 3^+} \frac{|x - 3|}{x - 3} = \lim_{x \to 3^+} \frac{x - 3}{x - 3} = 1 \]<br /><br />For \( x \to 3^- \), \( |x - 3| = -(x - 3) \):<br />\[ \lim_{x \to 3^-} \frac{|x - 3|}{x - 3} = \lim_{x \to 3^-} \frac{-(x - 3)}{x - 3} = -1 \]<br /><br />Since the left-hand limit and the right-hand limit are not equal, the two-sided limit does not exist.<br /><br />### Summary<br /><br />1. The derivative of \( y = 2x^2 + \ln(\sqrt{x}) \) is:<br />\[ \frac{dy}{dx} = 4x + \frac{2x} \]<br /><br />2. The limit \(\lim_{x \to 1} \frac{x^3 - 1}{x^2 - 1}\) is:<br />\[ \frac{3}{2} \]<br /><br />3. The limit \(\lim_{x \to 3} \frac{|x - 3|}{x - 3}\) does not exist because the left-hand limit and the right-hand limit are not equal.