Can someone please explain to me the steps to get the limit of the following? $\displaystyle \frac{x^4-1}{x-1}$ As $\displaystyle x \rightarrow 1$ Thanks.
Last edited by softwareguy; Mar 26th 2010 at 08:17 PM.
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Originally Posted by softwareguy Can someone please explain to me the steps to get the limit of the following? $\displaystyle x^4-1/x-1$ As X --> 1 Thanks. $\displaystyle \frac{x^{4}-1}{x-1}=\frac{(x^{2}+1)(x^{2}-1)}{x-1}=\frac{(x^{2}+1)(x+1)(x-1)}{x-1}=(x^{2}+1)(x+1)$ Now calculate the limit as x goes to 1
Last edited by harish21; Mar 22nd 2010 at 07:23 PM.
Somehow the answer is 4. How did they get the answer 4? I don't see it...
Originally Posted by softwareguy Somehow the answer is 4. How did they get the answer 4? I don't see it... Did you even try this? You are left with $\displaystyle (x^{2}+1)(x+1)$ As x goes to 1 (X--->1), the limit becomes: $\displaystyle (1^{2}+1)(1+1) = (2)(2)$ What do you get when you multiply 2 and 2?
Ouch, sorry about that! I totally spaced. Thanks very much for your help!
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