Limit Question

[Joe Mercurio]

Problem: Evaluate the following limit

Lim (x+sqrt(x)-2)
x->1 (x³-1)

I factored it into the following

[sqrt(x)+2][sqrt(x)-1] ---> I assumed that as X approaches 1, [sqrt(x)-1] = [x - 1] allowing me to cancel them out resulting in:
(x^2+x+1)(x-1)

[sqrt(x)+2] which would evaluate to 3/3 = 1, however from looking at the graph the limit is obviously 1/2. Not quite sure where I went wrong. Was the step above a false assumption? Did
[x^2+x+1] I not go far enough? Besides how I factored above, I'm not sure how else to reduce this equation.

 

[HallsofIvy]

Problem: Evaluate the following limit

Lim (x+sqrt(x)-2)
x->1 (x³-1)

I factored it into the following

[sqrt(x)+2][sqrt(x)-1] ---> I assumed that as X approaches 1, [sqrt(x)-1] = [x - 1]

There's your error. \(\displaystyle \sqrt{x}- 1\) is only equal to x at x= 1, not close to it. Instead use the fact that \(\displaystyle x- 1= (\sqrt{x}-1)(\sqrt{x}+1)\).

allowing me to cancel them out resulting in:
(x^2+x+1)(x-1)

[sqrt(x)+2] which would evaluate to 3/3 = 1, however from looking at the graph the limit is obviously 1/2. Not quite sure where I went wrong. Was the step above a false assumption? Did
[x^2+x+1] I not go far enough? Besides how I factored above, I'm not sure how else to reduce this equation.

 

[Joe Mercurio]

Instead use the fact that \(\displaystyle x- 1= (\sqrt{x}-1)(\sqrt{x}+1)\).

Much appreciated!

 

[lookagain]

Problem: Evaluate the following limit

Lim (x+sqrt(x)-2)
x->1 (x³-1)

I find all of this staying with the radicals to be relatively awkward and
taking many more characters to type out.


\(\displaystyle \displaystyle\lim_{x\to 1} \ \frac{x + \sqrt{x} - 2}{x^3 - 1}\)


\(\displaystyle Let \ \ y = \sqrt{x}. \ \ \ Then \ \ y^2 = x.\)


\(\displaystyle As \ \ {x\to 1}, \ \ {\sqrt{x}\to1}\) . . . . . . Why is this?


\(\displaystyle Then \ \ it \ \ would \ \ be \ \ the \ \ case \ \ that \ \ {y\to1} \) . . . . . . \(\displaystyle Recall \ \ that \ \ \sqrt{x} \ \ = \ \ y.\)


\(\displaystyle \displaystyle\lim_{y\to 1} \ \frac{y^2 + y - 2}{y^6 - 1} \ \ =\)


\(\displaystyle \displaystyle\lim_{y\to1} \ \frac{(y - 1)(y + 2)}{(y - 1)(y^5 + y^4 + y^3 + y^2 + y + 1)} \ \ = \ ?\)



Cancel the like factors between the numerator and the denominator.

Then continue evaluating the limit . . .

 

[Joe Mercurio]

JeffM,

Your breakdown was much appreciated, HallsofIvy gave me the right answer, but you explained where I went wrong, thanks again

 

[lookagain]

I assumed that as X approaches 1, [sqrt(x)-1] = [x - 1] ...

It is true that \(\displaystyle \displaystyle\lim_{x\to 0^+}\dfrac{\sqrt{x} - 1}{x - 1} \ = \ 1.\)


Your supposition made me think of this contrasting example.

 

[lookagain]

It is true that \(\displaystyle \displaystyle\lim_{x\to 0^+}\dfrac{\sqrt{x} - 1}{x - 1} \ = \ 1.\)


Your supposition made me think of this contrasting example.

My example is not an indeterminate form. I don't see a need for a
particular substitution, other than possibly these:


\(\displaystyle \displaystyle\lim_{x\to 0^+}\sqrt{x} \ = \ 0, \ \ \ as \ \ well \ \ as \ \ \ \displaystyle\lim_{x\to0^+} x \ = \ 0.\)


\(\displaystyle So, \ \ \displaystyle\lim_{x\to 0^+}\dfrac{\sqrt{x} - 1}{x - 1} \ = \)


\(\displaystyle \dfrac{0 - 1}{0 - 1} \ = \)


\(\displaystyle \dfrac{-1}{-1} \ = \)


\(\displaystyle \boxed{1}\)

 

[lookagain]

\(\displaystyle \text{OK If you say your substitution of } y\ for\ \sqrt{x} \text{ was not elegant, who am I to disagree.}\)

I am mistaken. I did not know you referred to the work in post # 5.

I thought you referred to some other subpart

Now I understand you were referring to the substitution of \(\displaystyle y \ \ for \ \ \sqrt{x}.\)