How to find the limit of a cubic conjugate?

Question

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->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!

Answer 1

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->0 and x+1=u then u is approaching 1 right?

Now we can use l'hospital.

lim u->1 = 0/0

take derivative of the top and bottom.

lim u ->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

Answer 2

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-->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

Answer 3

Here's another substitution technique
similar to xian gaon's answer.

lim (x -> 0) [ 1 - (x + 1)^(1/3) ] / x

Let u = 1 - (x + 1)^(1/3)
First, notice here, that as x -> 0, u -> 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 -> 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 -> 0) 1 / ( -3 + 3u - u^2 )

Now we see that as u -> 0, the limit is :

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

Answer 4

there are various consumer-friendly consumer-friendly strategies to locate the shrink. Plug interior the value that x techniques into the shrink, wherein case we get 0 / 0. in case you get 0 / 0, you are able to, carry out a little algebraic manipulation; l'hopital's rule; attempt u - substitution; and much extra to get the respond. (a million / 2)(0 + a million)^(-a million / 2) = a million / 2 (a million / 2)(9)^(-a million / 2) = a million / 6

Answer 5

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)