Use numerical and graphical evidence to guess the value of the limit $\underset{x \to 1} \lim \frac{x^3-1}{\sqrt{x}-1}$? 
Use numerical and graphical evidence to guess the value of the limit $$\underset{x \to 1} \lim \frac{x^3-1}{\sqrt{x}-1}$$

I am a Calculus 1 student, and I'm not sure what this problem in my textbook means when it says, "use numerical and graphical evidence". I worked around with it a bit and found that the answer is 6. However, I'm not sure if this is what my professor wants. 
The textbook is "Calculus: Early Transcendentals", 8th Edition, by James Stewart. This is problem 55a in section 2.2.
What does this question mean by, "Use numerical and graphical evidence"?
$$\underset{x \to 1} \lim \frac{x^3-1}{\sqrt{x}-1}$$
$$=\underset{x \to 1} \lim \frac{x^3-1}{\sqrt{x}-1}\cdot \frac{\sqrt{x} + 1}{\sqrt{x} + 1}$$
$$=\underset{x \to 1} \lim \frac{(x^3-1)(\sqrt{x} + 1)}{x-1}$$
$$=\underset{x \to 1} \lim \frac{(x-1)(x^2+x+1)(\sqrt{x} + 1)}{x-1}$$
$$=\underset{x \to 1} \lim \;\ (x^2+x+1)(\sqrt{x} + 1)$$
$$= ((1)^2+(1)+1)(\sqrt{1}+1)=(3)(2)=6$$
 A: To guess the limit, see this graph:
 
A: You did it well but it could be done faster.
Let $x=y^2$
$$\underset{x \to 1} \lim \frac{x^3-1}{\sqrt{x}-1}=\underset{y \to 1} \lim \frac{y^6-1}{y-1}=1+y+y^2+y^3+y^4+y^5\,\, \to 6$$
If you want to also see how the limit is approached, let $x=1+t$
$$\underset{x \to 1} \lim \frac{x^3-1}{\sqrt{x}-1}=\underset{t \to 0} \lim \frac{(1+t)^3-1}{\sqrt{1+t}-1}$$ 
Using the binomial expansion or Taylor series
$$\frac{(1+t)^3-1}{\sqrt{1+t}-1}=\frac{3t+3t^2+t^3 } {\left( 1+\frac{t}{2}-\frac{t^2}{8}+\frac{t^3}{16}+O\left(t^4\right)\right)-1 }=6+\frac{15 t}{2}+O\left(t^2\right)$$ that is to say
$$\frac{x^3-1}{\sqrt{x}-1}=6+\frac{15}{2}(x-1)+O\left((x-1)^2\right)$$
Using the examples given by D.B., you get $6+\frac{15}{2}(0.99-1)=5.925$ and $6+\frac{15}{2}(1.01-1)=6.075$.
A: To try to find the value of the limit numerically, you just need to try values of $x$ close to $1$.  To be thorough, you might want to check above and below $1$.  For example,
$$\frac{(1.01)^3-1}{\sqrt{1.01}-1} = 6.08.$$
$$\frac{(0.99)^3-1}{\sqrt{0.99}-1} = 5.93.$$
A: Great job. 
Most likely, the book is trying to guide beginners to the concept of limit. The exercise could be too easy for you.
The book expect you to try out a few values around $x=1$ and sketch out a plot. 
import math

def f(x):
    return (x**3-1)/(math.sqrt(x)-1)

for i in range(5):
    x = 1 + 10**(-i)
    print(x,f(x))
for i in range(5):
    x = 1 - 10**(-i)
    print(x, f(x))

produces 
(2, 16.8994949366)
(1.1, 6.78155728744)
(1.01, 6.07531281196)
(1.001, 6.00750312531)
(1.0001, 6.00075003126)
(0, 1.0)
(0.9, 5.28093173772)
(0.99, 5.92531218695)
(0.999, 5.99250312469)
(0.9999, 5.99925003125)

If you collect sufficient point, you can draw a sketch of the picture.
