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Find the value of b, b ≥ 0, for which limx->0 {(sqrt(3x+b)-5)/x} exists
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\[\lim_{x \rightarrow 0}\frac{\sqrt{3x+b}-5}{x} \cdot \frac{\sqrt{3x+b}+5}{\sqrt{3x+b}+5}\] \[\lim_{x \rightarrow 0}\frac{(3x+b)-25}{x(\sqrt{3x+b}+5)}\] so we want the x on top to cancel x on bottom only way this happens if we let b-25=0 => b=25
\[\lim_{x \rightarrow 0}\frac{3x}{x(\sqrt{3x+25}+5)}=\lim_{x \rightarrow 0}\frac{3}{\sqrt{3x+25}+5}=\frac{3}{\sqrt{3(0)+25}+5}=\frac{3}{10}\] see limit exists! :)
THANK YOU! :)
np
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