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Calc 1 Limit Problem No L'hopital's rule allowed! Posting limit
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\[\lim_{x \rightarrow 1}\frac{ \sqrt{1+3x}-2 }{ \sqrt{x+8}-3 }\]
Multiplying by the conjugates didn't quite work out, did it?
Nope. Neither top nor bottom.
@nincompoop will help, reeezun: cuz b0red.kthxbye.
\(\color{blue}{\text{Originally Posted by}}\) @SithsAndGiggles Worked for me: \[\frac{\sqrt{3x+1}-2}{\sqrt{x+8}-3}\cdot\frac{\sqrt{3x+1}+2}{\sqrt{3x+1}+2}\cdot\frac{\sqrt{x+8}+3}{\sqrt{x+8}+3}\] yields \[\frac{3x+1-4}{x-1}\cdot\frac{\sqrt{x+8}+3}{\sqrt{3x+1}+2}\] \(\color{blue}{\text{End of Quote}}\)
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Ah, both conjugates, not just one. Duh x_x
Yup, works out nicely after that.
Alrighty, cool, thanks :3
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