Doubt for substitution in two variables limit I have an exercise on my book like that
$\lim_{(x,y)\to (0,1)}f(x,y)=\frac {(y^2-1)(9x^2+2)\log\left(1+x^5\right)}{x^4+(y-1)^6}$
then it says without other explanations:
for $(x,y)\to(0,1)$ you have 
$\left\vert f(x,y)\right\vert \le 5\left\vert\frac{(y-1)x^5}{x^4}\right\vert=5\left\vert(y-1)x\right\vert \to 0$
Shame on me maybe, but I really can't see how 
${\left(y^2-1\right)\left(9x^2+2\right)}$ becomes that $5\vert(y-1)\vert$
Thank you as always.
 A: When estimating the absolute value of a fraction, you needn't limit yourself to algebraic equalities, but may also make the (absolute value of) the numerator larger and the denominator smaller. If $x \neq 0$, for example,
\begin{align*}
\left\lvert\frac{(y^{2} - 1) (9x^{2}+ 2) \log(1 + x^{5})}{x^{4} + (y - 1)^{6}}\right\rvert
  &\leq \left\lvert\frac{(y^{2} - 1) (9x^{2}+ 2) \log(1 + x^{5})}{x^{4}}\right\rvert && \text{shrink denominator} \\
  &= \left\lvert\frac{(y - 1)(y + 1) (9x^{2}+ 2) \log(1 + x^{5})}{x^{4}}\right\rvert && \text{difference of squares} \\
  &\leq \left\lvert\frac{4.5(y - 1) \log(1 + x^{5})}{x^{4}}\right\rvert && \begin{gathered}y + 1 \approx 2,\\ 9x^{2} + 2 \approx 2\end{gathered} \\
  &\leq \left\lvert\frac{5(y - 1) x^{5}}{x^{4}}\right\rvert && \frac{\log(1 + x^{5})}{x^{5}} \approx 1.
\end{align*}
The last two steps are where most of the work occurs. Thanks to the formal definition of a limit, the left-hand side of each approximation does not exceed a number slightly larger than the right-hand side, provided $(x, y)$ is sufficiently close to $(0, 1)$. The constant $4.5$ is just a convenient number larger than $2 \times 2 = 4$, and the ratio $5/4.5$ in the last inequality is a convenient number larger than $1$.
(If $x = 0$, the fraction is $0$ for all $y \neq 1$, so there's nothing to prove.)
