(sqrt7-x)/(49-x^4) as x approaches sqrt7 I multiplied by the conjugate to rationalize the numerator. I also factored the bottom... and I came up with this. (7-x)/(x^2+7)(x^2-7) What to do after this?
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$\displaystyle \lim_{x \to \sqrt{7}} \frac{\sqrt{7} - x}{49-x^4}$ $\displaystyle = \lim_{x \to \sqrt{7}} \frac{\sqrt{7} - x}{{\color{red}(7-x^2)}(7+x^2)}$ Try applying the difference of squares formula to the red
haha so obvious... sometimes I don't see these things. Thank you sir!
Originally Posted by Sturm88 Hey I have a really silly question, but what is a root + a root. like sqrt7 + sqrt 7 in the problem I just posted... $\displaystyle a + a = 2a$. This is obvious right? Imagine $\displaystyle a = \sqrt{7}$
So the answer is 28sqrt7?
(1/64)-(1/x^2)/(x-8) as x approaches 8 I used difference of squares on the top and I got this (1/8x+x)(1/8-x)/(x-8) stuck here.
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