<s>[INST] 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"?
$$\underset{x \to 1} \lim \frac{x^3-1}{\sqrt{x}-1}$$
$$=\underset{x \to 1} \lim \frac{x^3-1}{\sqrt{x}-1}\cdot \frac{\sqrt{x} + 1}{\sqrt{x} + 1}$$
$$=\underset{x \to 1} \lim \frac{(x^3-1)(\sqrt{x} + 1)}{x-1}$$
$$=\underset{x \to 1} \lim \frac{(x-1)(x^2+x+1)(\sqrt{x} + 1)}{x-1}$$
$$=\underset{x \to 1} \lim \;\ (x^2+x+1)(\sqrt{x} + 1)$$
$$= ((1)^2+(1)+1)(\sqrt{1}+1)=(3)(2)=6$$
 [/INST] 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 $6+\frac{15}{2}(1.01-1)=6.075$.
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