Substitution and Cancelling the Common Factors - Giri-Subrahmanya/temp GitHub Wiki

In this section, we shall take a look at two approaches to find the limit of a function.

  1. Substitution
  2. Cancelling the Common Factors

SUBSTITUTE

A straightforward method to find the limit of a function at a specific point is to substitute the point into the function.

EXAMPLE-1

$\mathbb{Find}\lim_{(x,y)\rightarrow (2,3)}\frac{x^{2}+y^{2}-y}{y^{2}-x^{2}}$.

$$ \begin{matrix} \lim_{(x,y)\rightarrow (2,3)}\frac{x^{2}+y^{2}-y}{y^{2}-x^{2}} & = & \frac{(2)^{2}+(3)^{2}-3}{(3)^{2}-(2)^{2}} \ & & \ & =& \frac{4+9-3}{9-4} \ & & \ & =& \frac{10}{5} \ & & \ & =& 2 \end{matrix} $$

EXAMPLE-2

$\mathbb{Given}\ f(x,y)=\frac{x^{2}+y^{2}+y(y-2x-1)}{x^{2}-y^{2}}$ and $g(x,y)=\frac{2y^{2}+x(x-2y-1)}{x^{2}-y^{2}}$.

$\mathbb{Find}\lim_{(x,y)\rightarrow (1,1)}(f(x,y)-g(x,y))$

Substituting $\displaystyle ( 1,1)$ into $\displaystyle f( x,y)$ and $\displaystyle g( x,y)$ would not be possible. However, substituting $\displaystyle ( 1,1)$ into $\displaystyle f( x,y) -g( x,y)$ is possible.

$$ \begin{matrix} f(x,y)-g(x,y) & =& \frac{x^{2}+y^{2}+y(y-2x-1)}{x^{2}-y^{2}}-\frac{2y^{2}+x(x-2y-1)}{x^{2}-y^{2}} \ & & \ & = & \left(\frac{x^{2}+y^{2}+y^{2}-2xy-y)}{x^{2}-y^{2}}\right)-\left(\frac{2y^{2}+x^{2}-2xy-x}{x^{2}-y^{2}}\right) \ & & \ & = & \frac{x^{2}+2y^{2}-2xy-y-2y^{2}-x^{2}+2xy+x}{x^{2}-y^{2}} \ & & \ & = & \frac{x-y}{x^{2}-y^{2}} \ & & \ & = & \frac{x-y}{(x-y)(x+y)}\ & & \ & = & \frac{1}{x+y} \end{matrix}$$

Now, we can substitute and find the limit.

$$ \begin{matrix} \lim_{(x,y)\rightarrow (1,1)}(f(x,y)-g(x,y)) & = & \lim_{(x,y)\rightarrow (1,1)}\left(\frac{1}{x+y}\right)\ & & \ & = & \frac{1}{1+1}\ & & \ & = & 0.5 \end{matrix}$$

CANCEL THE COMMON FACTORS

When we try to substitute the point into the function, the denominator might become \math-container{0}. This can sometimes be tackled by canceling the common factors from both the numerator and the denominator.

A few algebraic identities might be needed. Some are given below:

  • $(a+b)^{2} =a^{2} +2ab+b^{2}$
  • $(a-b)^{2} =a^{2} -2ab+b^{2}$
  • $a^{2} -b^{2} =( a+b)( a-b)$
  • $( x-a)( x-b) =x^{2} -x( a+b) +ab$
  • $(a+b)^{3} =a^{3} +3a^{2} b+3ab^{2} +b^{3} =a^{3} +b^{3} +3ab( a+b)$
  • $(a-b)^{3} =a^{3} -3a^{2} b+3ab^{2} +b^{3} =a^{3} +b^{3} -3ab( a-b)$
  • $a^{3} +b^{3} +c^{3} -3abc=( a+b+c)\left( a^{2} +b^{2} +c^{2} -ab-bc-ac\right)$
  • $(a+b+c)^{2} =a^{2} +b^{2} +c^{2} +2ab+2ac+2bc$

EXAMPLE-3

$\mathbb{Find} \displaystyle \lim _{( x,y)\rightarrow ( 3,1)}\left(\frac{x^{2} -3xy^{2}}{x^{2} -9y^{4}}\right)$

If we substitute $(3,1)$ into the denominator, we will get $0$. Let us cancel the common factors.

$$ \begin{matrix} \lim _{( x,y)\rightarrow ( 3,1)}\left(\frac{x^{2} -3xy^{2}}{x^{2} -9y^{4}}\right) & = & \lim _{( x,y)\rightarrow ( 3,1)}\left(\frac{x\left( x-3y^{2}\right)}{\left( x-3y^{2}\right)\left( x+3y^{2}\right)}\right)\ & & \ & = & \lim _{( x,y)\rightarrow ( 3,1)}\left(\frac{x}{x+3y^{2}}\right) \end{matrix} $$

Now, substitution is possible.

$$ \begin{matrix} \lim _{( x,y)\rightarrow ( 3,1)}\left(\frac{x}{x+3y^{2}}\right) & = &\frac{3}{3+3( 1)^{2}}\ & & \ & =& \frac{3}{3+3}\ & & \ & =& 0.5 \end{matrix} $$