# Find $\lim_{x \rightarrow 1} \frac{\sqrt{2x^2-1}-1}{x^3-1}$ without using L'Hôpital's rule

I tried:

$$\lim_{x \rightarrow 1} \frac{\sqrt{2x^2-1}-1}{x^3-1} = \\ \lim_{x \rightarrow 1} \frac{\sqrt{2x^2-1}-1}{x^3-1} \cdot \frac{\sqrt{2x^2-1}+1}{\sqrt{2x^2-1}+1}\\ = \lim_{x \rightarrow 1} \frac{4x^4-4x^2}{(x^3-1)(\sqrt{2x^2-1}+1)} = ???$$

My questions:

1. What do I do next? Am I doing it correctly so far?
2. How do you know which method to use when simplifying expressions? How do you know whether if you have to rationalize the denominator or the numerator or both, divide the numerator and the denominator by one or the other, factorize, use more than just one method, etc...? I find myself wasting a lot of time moving from one method to the other never really knowing why I am using a certain method or whether if it works or not. Do you know any tricks or have any advice for me?
• You shouldn't write $\displaystyle{\lim_{x\to1}\left(\cdots\right)=\lim_{x\to1}\left(\cdots\right)=\cdots=}$, but rather transform the expression whose limit you are looking for, until you reach an appropriate form, and then take the limit ! – Adren Feb 18 '17 at 16:38
• For a very general trick, see math.stackexchange.com/a/2131837/21820 and math.stackexchange.com/a/2122660/21820. Namely, use asymptotic expansion. Even if you do not know the limit, this will let you find it. And even if you cannot use this method in an exam, using it will let you see what are the individual pieces to resolve by elementary means as shown in the first link. – user21820 Feb 19 '17 at 2:12

There is an error in the third line which is overlooked.

$$\lim_{x \rightarrow 1} \frac{\sqrt{2x^2-1}-1}{x^3-1} = \\ \lim_{x \rightarrow 1} \frac{\sqrt{2x^2-1}-1}{x^3-1} \cdot \frac{\sqrt{2x^2-1}+1}{\sqrt{2x^2-1}+1}\\ = \lim_{x \rightarrow 1} \frac{2x^2-2}{(x^3-1)(\sqrt{2x^2-1}+1)}$$

$$=\lim_{x \rightarrow 1} \frac{2(x-1)(x+1)}{(x-1)(x^2+x+1)(\sqrt{2x^2-1}+1)}$$

$$=\lim_{x \rightarrow 1} \frac{2(x+1)}{(x^2+x+1)(\sqrt{2x^2-1}+1)} = \frac {2}{3}$$

1) You made a small error:

$$\frac{(\sqrt{2x^2-1}-1)(\sqrt{2x^2-1}+1)}{(x^3-1)(\sqrt{2x^2-1}+1)}=\frac{(2x^2-1)-1^2}{(x-1)(x^2+x+1)(\sqrt{2x^2-1}+1)}$$

Note the numerator should be the difference of two squares. Factor out $(x-1)$ from the numerator, and then you may now directly substitute $x=1$.

2) In general, try to spot out which parts of the limit cause the $0/0$ indeterminate form part. Those clearly are factorable, so remove the right factors from them.

HINT: use that $$x^3-1=(x-1)(x^2+x+1)$$ and $$4x^4-4x^2=4x^2(x-1)(x+1)$$ can you finish now? and use that $$2x^2-1-1=2x^2-2=2(x^2-1)=2(x-1)(x+1)$$

• :-) Nicely factored, and full reading of the entire question. – Simply Beautiful Art Feb 18 '17 at 16:26
• Actually, we are all goofs. See my updated answer – Simply Beautiful Art Feb 18 '17 at 16:40
• How did you know that $x^3-1=(x-1)(x^2+x+1)$? – Mark Read Feb 18 '17 at 16:55
• @MarkRead Finite geometric series: $$1+x+x^2+\dots+x^{n-1}=\frac{x^n-1}{x-1}$$ – Simply Beautiful Art Feb 18 '17 at 16:55
• note that $$\frac{x^3-1}{x-1}=x^2+x+1$$ for $$x\ne 1$$ – Dr. Sonnhard Graubner Feb 18 '17 at 17:09

Your last step seems incorrect because you didn't multiply the two numerators correctly.

Continue your calculation as follows. Your limit is $$\lim_{x \to 1} \frac{2x^2 - 2}{(x^3 - 1)(\sqrt{2x^2 - 1} + 1)} = \lim_{x \to 1} \frac{2}{\sqrt{2x^2 - 1} + 1} \cdot \lim_{x \to 1}\frac{x^2 - 1}{x^3 - 1} = 1 \cdot \lim_{x \to 1}\frac{x^2 - 1}{x^3 - 1}.$$

Before going any further, let me point out a couple of things.

First, when you're trying to calculate a limit of a quantity expressed as a product or quotient of several factors, any factors that have a finite, nonzero limit are easy to deal with: just split them off as we've done above.

Second, the method of rationalization is most often applied when you have a complicated term $A \pm B$ whose limit is zero or infinity, or unknown, but the expression $A^2 - B^2$ is simpler to deal with. This is useful when the "conjugate" expression $A \mp B$ has a finite nonzero limit.

In this case, the expression $A - B = \sqrt{2x^2 - 1} - 1$ had a zero limit and was therefore difficult to deal with. However, the conjugate term $A + B = \sqrt{2x^2 - 1} + 1$ had a limit of $2\ne 0, \infty$, and the product $(A-B)(A + B) = A^2 - B^2 = 2x^2 - 2$ is simple in form.

To continue the calculation, there are two methods, depending on whether it's easy for you to factor the numerator and denominator.

If it is easy to factor, do this: $$\lim_{x \to 1} \frac{x^2 - 1}{x^3 - 1} = \lim_{x \to 1} \frac{(x-1)(x +1)}{(x - 1)(x^2 + x + 1)} = \lim_{x \to 1} \frac{x + 1}{x^2 + x + 1} = \frac{2}{3}$$

Otherwise, make the substitution $x = 1 + h$, where $h \to 0$: $$\lim_{x \to 1} \frac{x^2 - 1}{x^3 - 1} = \lim_{h \to 0} \frac{(1 + h)^2 - 1}{(1 + h)^3 - 1} = \lim_{h \to 0} \frac{2h + h^2}{3h + 3h^2 + h^3} = \lim_{h \to 0} \frac{h(2 + h)}{h(3 + 3h + h^2)} = \lim_{h \to 0} \frac{2 + h}{3 + 3h + h^2} = \frac{2}{3}$$

Since $x^3-1$ is a polynomial that is $0$ at $x=1,$ we can factor an $(x-1)$ out: $x^3-1 = (x-1)(x^2+x+1).$ Our expression thus equals

$$\tag 1 \frac{\sqrt {2x^2-1} - 1}{x-1}\frac{1}{x^2 + x + 1}.$$

Let $f(x) = \sqrt {2x^2-1}.$ We can write $(1)$ as

$$\tag 2 \frac{f(x) - f(1)}{x-1}\frac{1}{x^2 + x + 1}.$$

As $x\to 1,$ the first fraction in $(2) \to f'(1)$ by definition of the derivative. The second fraction $\to 1/3.$ The desired limit is thus $f'(1)/3,$ which is a simple computation. (And no L'Hopital was used.)