What is the value of the limit
limx--%26gt;-鈭? 3鈭?x虏-1)/x-1|||lim sqrt ( x^2 - 1) / (x - 1) , ( x goes to infinity )
use L ' hospital ' rule
first limit
= lim[ 3(1/2)(x^2 - 1) ^ ( - 1/ 2) 2x] / 1 , ( x goes infinity )
= infinity|||2.13|||raise (x2-1) to the 1/3 power (instead of using the cube root) and u will get the answer.|||Since x is approaching negative infinity, the 1's become negligible. Essentially we are dealing with the limit of 3x/x as x approaches negative infinity. But since the x in the numerator is squared, it is positive. So the limit is -3.|||Bring everything up to square
limx--%26gt;-鈭?9(x虏 - 1) / (x - 1)虏
limx--%26gt;-鈭?(9x虏 - 9) / x虏 - 2x +1
Now we can take the squared and neglect everything else
limx--%26gt;-鈭?9x虏 / x虏 = 9
but we need to root it back
鈭? = 3
Altho putting it up to square messes with negative values so the answer might be -3.
so limx--%26gt;-鈭?3鈭?x虏-1) / (x-1) = 3|||I am assuming that you mean 3鈭?x虏-1)/(x-1)
3鈭?x虏-1)/(x-1) = 3鈭歔(x + 1)(x - 1)]/(x - 1) = 3鈭?x + 1) * [鈭歔(x - 1)]/(x - 1) =
limx--%26gt;-鈭?| 3鈭歔(x + 1)/(x - 1)] | = 3
limx--%26gt;-鈭?3鈭?x虏-1)/x-1 = -3
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