Calculating Limit of Function – Difference of rational functions to one – Exercise 6311

Exercise

Evaluate the following limit:

$\lim _ { x \rightarrow 1} (\frac{3}{1-x^3}-\frac{2}{1-x^2})$

Final Answer

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Solution

First, we try to plug in $x = 1$ and get

$\frac{3}{1-1^3}-\frac{2}{1-1^2}=$

$\frac{3}{0}-\frac{2}{0}=$

$=\infty-\infty$

Note: The zero in the denominator is not absolute zero, but a number that is tending to zero.

We got the phrase $\infty-\infty$ (=infinity minus infinity). This is an indeterminate form, therefore we have to get out of this situation.

$\lim _ { x \rightarrow 1} (\frac{3}{1-x^3}-\frac{2}{1-x^2})=$

In order to use Lopital Rule, we will break down the polynomials into factors and calculate the least common denominator

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

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

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

$=\lim _ { x \rightarrow 1} \frac{-2x^2+x+1}{1+x+x^2-x^2-x^3-x^4}=$

$=\lim _ { x \rightarrow 1} \frac{-2x^2+x+1}{1+x-x^3-x^4}=$

We plug in one again and get

$= \frac{-2\cdot 1^2+1+1}{1+1-1^3-1^4}=$

$= \frac{0}{0}$

We got the phrase $\frac{"0"}{"0"}$ (=a number tending to zero divides by a number tending to zero). This is also an indeterminate form, in such cases we use Lopital Rule – we derive the numerator and denominator separately and we get

$=\lim _ { x \rightarrow 1} \frac{-4x+1}{1-3x^2-4x^3}=$

We plug in one again and this time we get

$=\frac{-4\cdot 1+1}{1-3\cdot 1^2-4\cdot 1^3}=$

$=\frac{-3}{-6}=$

$=\frac{1}{2}$

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