The values of a, b, c for which the function | vedclass

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(D) For the function to be continuous at $x = 0$, we must have $\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0^+} f(x) = f(0)$.First, calculate the left-hand l

Vedclass · Accepted Answer

$a = -2, b \in \mathbb{R} - \{0\}, c = 0$

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(D) For the function to be continuous at $x = 0$, we must have $\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0^+} f(x) = f(0)$.First, calculate the left-hand limit $(LHL)$:$\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0^-} \frac{\sin(a+1)x + \sin x}{x} = \lim_{x 	o 0^-} \left( \frac{\sin(a+1)x}{x} + \frac{\sin x}{x} ight) = (a+1) + 1 = a+2$.Next, calculate the right-hand limit $(RHL)$:$\lim_{x 	o 0^+} f(x) = \lim_{x 	o 0^+} \frac{\sqrt{x+bx^2} - \sqrt{x}}{bx^{1/2}} = \lim_{x 	o 0^+} \frac{\sqrt{x}(\sqrt{1+bx} - 1)}{bx^{1/2}} = \lim_{x 	o 0^+} \frac{\sqrt{1+bx} - 1}{b}$.Using the binomial expansion $(1+bx)^{1/2} \approx 1 + \frac{1}{2}bx$, we get:$\lim_{x 	o 0^+} \frac{1 + \frac{1}{2}bx - 1}{b} = \lim_{x 	o 0^+} \frac{\frac{1}{2}bx}{b} = \lim_{x 	o 0^+} \frac{x}{2} = 0$.Since $f(0) = c$, for continuity we require $a+2 = 0$ and $c = 0$.Thus, $a = -2$ and $c = 0$. Since the denominator in the $RHL$ expression contains $b$, we must have $b 
eq 0$ for the function to be defined for $x > 0$.Therefore, $a = -2, b \in \mathbb{R} - \{0\}, c = 0$.

The values of $a, b, c$ for which the function $f(x) = \begin{cases} \frac{\sin(a+1)x + \sin x}{x}, & x < 0 \\ c, & x = 0 \\ \frac{(x+bx^2)^{1/2} - \sqrt{x}}{bx^{1/2}}, & x > 0 \end{cases}$ is continuous at $x = 0$, are

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