Define f(x) = begin{cases} b - ax & text{if } | vedclass

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(B) For the limit $\lim_{x \rightarrow 2} f(x)$ to exist,the left-hand limit $(LHL)$ must equal the right-hand limit $(RHL)$.$LHL = \lim_{x \rightarro

Vedclass · Accepted Answer

$-1$

Vedclass · Answer

(B) For the limit $\lim_{x ightarrow 2} f(x)$ to exist,the left-hand limit $(LHL)$ must equal the right-hand limit $(RHL)$.$LHL = \lim_{x ightarrow 2^-} f(x) = \lim_{x ightarrow 2^-} (b - ax) = b - 2a$.$RHL = \lim_{x ightarrow 2^+} f(x) = \lim_{x ightarrow 2^+} (a + 2bx) = a + 4b$.Since the limit exists,$b - 2a = a + 4b$.Rearranging the terms: $-2a - a = 4b - b$,which simplifies to $-3a = 3b$.Dividing both sides by $3b$ (assuming $b 
eq 0$),we get $\frac{a}{b} = -1$.

Define $f(x) = \begin{cases} b - ax & \text{if } x < 2 \\ 3 & \text{if } x = 2 \\ a + 2bx & \text{if } x > 2 \end{cases}$. If $\lim_{x \rightarrow 2} f(x)$ exists,then find the value of $\frac{a}{b}$.

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