Limit !!!!!!!!!!!!!!?

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