How to find the limit of a cubic conjugate?

Hey,





I have this one calculus problem that asks to evaluate the following limit in other words find the limit.





The question is:





lim {1 - [ cuberoot(x+1) ] } / x


x-%26gt;0.





How would I solve this, I have worked with complex conjugates before but they were just square root not cubic root. Is there some certain way or can I just apply the same concept of multiplying by complex conjugate?





Thanks!|||This was a bit tricky but i think i got it.





The main problem was it wasn't of indeterminate form so we couldn't use l'hostipal.





Make the sub u^3 = (x+1)





then





u^3 - 1 = x





and now sub in terms





(1-cuberoot (u^3))/(u^3 - 1)





or





(1-u)/(u^3-1)





we also need to change the what the limit is approaching.





so as x-%26gt;0 and x+1=u then u is approaching 1 right?





Now we can use l'hospital.





lim u-%26gt;1 = 0/0





take derivative of the top and bottom.





lim u -%26gt;1 of ...





-1/(3*u^2)





so the answer is -1/3





I checked on the calculator to make sure this was correct. It is. Also i want to add that if you don't know l'hospital's rule you can also factor the bottom using the difference between cubes.





The equation becomes:





(1-u)/((u-1)*(u^2+u+1)) and then it is easy|||if you have a graphing calculator i would suggest that





i don't have one on me, but perhaps its 1/3 (i can't be sure without a graphing calculator)|||Remeber try to cancel out the top:





Remember the exponent laws: why is it that





square root(x) * square root(x)=x





this is because (x^a)(x^b)=x^a+b





x^1/2 * x^1/2= x^1/2+1/2=x^1





in this same manner we will try to cancel exponent 1/3





As such 1/3 +b =1 b=3/3 -1/3=2/3





(x+1)^1/3 -1/x * (x+1)^2/3 +1/(x+1)^2/3 +1=





x+1 -1/(x)(x+1)^2/3 the Xs cancel out





therefore 1/(x+1)^2/3 now place in the zero of the limit which equals =1








therefore limit x--%26gt;0 (x+1)^1/2 -1/x = 1











INDEED KNOWING AND USING THE EXPONENT LAWS YOU CAN FIND THE CONJUGATE TO ANY CUBIC ROOT:





SUCH AS LETS SAY THAT YOU WANT TO FIND THE CONGUTE OF A A CUBIC ROOT TO THE 243.





THE QUESTION Is limit approaching x to 2





where f(x)= (x-2)^1/243/x -2





1/243 +b =1 b= 243/243(since it is still one) -1/243= 242/243





(x-2)^1/243/x-1 * (x-2)^242/243/(x-2)^242/243





= x-2/(x-2)(x-2)^242/243





= 1/(x-2)^242/243= infinity|||Here's another substitution technique


similar to xian gaon's answer.





lim (x -%26gt; 0) [ 1 - (x + 1)^(1/3) ] / x





Let u = 1 - (x + 1)^(1/3)


First, notice here, that as x -%26gt; 0, u -%26gt; 0 also.





Rearrange :


(x + 1)^(1/3) = 1 - u





Cube both sides :


x + 1 = (1 - u)^3





Rearrange :


x = (1 - u)^3 - 1





Simplify :


x = -3u + 3u^2 - u^3





Now substitute u for [ 1 - (x + 1)^(1/3) ]


and ( -3u + 3u^2 - u^3 ) for x into the


original expression. This gives :





lim (u -%26gt; 0) u / ( -3u + 3u^2 - u^3 )





Because u is not zero, we can divide


both numerator and denominator by u.


The expression then becomes :





lim (u -%26gt; 0) 1 / ( -3 + 3u - u^2 )





Now we see that as u -%26gt; 0, the limit is :





1 / ( -3 + 0 - 0 ) = -1 / 3.