How to solve this limit (without using L'Hospital)?

Question

$$\lim_{x\to\sqrt{2}}\frac{x^9-3x^8+x^6-9x^4-4x^2-16x+84}{x^5-3x^4-4x+12}$$

As $x-\sqrt 2$ is a factor of both these expressions, so I guess one way is long division. But that would go so long and messy, isn't there any manipulation than can work?

Answer 1

If $\sqrt2$ is a root of a polynomial with rational coefficients, then that polynomial is a multiple of $\require{cancel}x^2-2$. In your case, you have\begin{multline}\lim_{x\to\sqrt2}\frac{x^9-3x^8+x^6-9x^4-4x^2-16x+84}{x^5-3x^4-4x+12}=\\=\lim_{x\to\sqrt2}\frac{\cancel{(x^2-2)}(x^7-3 x^6+2 x^5-5 x^4+4 x^3-19 x^2+8 x-42)}{\cancel{(x^2-2)}(x^3-3 x^2+2 x-6)}.\end{multline}Can you take it from here?

Answer 2

As an alternative, to avoid long division, we can let $u=x-\sqrt 2 \to 0$ and since $a_0u^0$ terms cancel out we can only keep the terms $a_1u$ to obtain by $x=u+\sqrt 2$

$$\frac{x^9-3x^8+x^6-9x^4-4x^2-16x+84}{x^5-3x^4-4x+12}=$$

$$=\frac{9(\sqrt 2)^8u-24(\sqrt 2)^7u+6(\sqrt 2)^5u-36(\sqrt 2)^3u-8(\sqrt 2)u-16u+O(u^2)}{5(\sqrt 2)^4u-12(\sqrt 2)^3u-4u+O(u^2)}=$$

$$=\frac{9(\sqrt 2)^8-24(\sqrt 2)^7+6(\sqrt 2)^5-36(\sqrt 2)^3-8(\sqrt 2)-16+O(u)}{5(\sqrt 2)^4-12(\sqrt 2)^3-4+O(u)}\to \frac{248 \sqrt 2-128}{24 \sqrt 2-16}$$

Answer 3

The remark that $\sqrt{2}$ is a root of a polynomial with integer coefficients can be extended further than just saying $x^2-2$ is a factor.

Because $\sqrt{2}$ is a surd, we can actually separate odd and even powers of $x$.

Notice that for denominator $\begin{cases}x^5-4x=x(x^4-4)\\-3x^4+12=-3(x^4-4)\end{cases}$ and so it factorizes to $(x-3)(x^4-4)$

For the numerator the odd part $x^9-16x=x(x^8-16)=x(x^4-4)(x^4+4)$ also has $(x^4-4)$ factor, so we can hope that the whole polynomial is factorizable in this way.

Doing the division you'll find $(x^4-4)(x^5+x^2+4x-3x^4-21)$ and the fact that the constant term is $21$ is very interesting, it means that divisors of $21$ may be roots to.

Since we have an $(x-3)$ factor in denominator, we are tempted to see if $3$ is also a root of numerator, which is indeed the case.

In the end you can simplify with $(x-3)(x^4-4)$ and there is no denominator left at all...