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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


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

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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