(e^(2x)-1)/(e^x-1) as x approaches 0
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Originally Posted by ihatemath (e^(2x)-1)/(e^x-1) as x approaches 0 $\displaystyle \frac{e^{2x} - 1}{e^x - 1} = \frac{(e^x + 1)(e^x - 1)}{e^x - 1}$ $\displaystyle = e^x + 1$. So what do you think this tends to as $\displaystyle x \to 0$?
oh my gosh! I can't believe I missed something so simple! Thanks!
$\displaystyle lim_{x \rightarrow 0} ~\frac{e^{2x} - 1}{e^x - 1} = lim_{x \rightarrow 0} ~\frac{(e^x - 1)(e^x + 1)}{e^x - 1} = lim_{x \rightarrow 0} ~(e^x + 1) = 2$
Originally Posted by dedust $\displaystyle lim_{x \rightarrow 0} ~\frac{e^{2x} - 1}{e^x - 1} = lim_{x \rightarrow 0} ~\frac{(e^x - 1)(e^x + 1)}{e^x - 1} = lim_{x \rightarrow 0} ~(e^x + 1) = 2$ You're a little late...
Originally Posted by Prove It You're a little late... yeah,..still newbie in LaTex
Don't you hate it when someone posts while you are writing a response?
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