Exercise
Evaluate the following limit:
\lim _ { x \rightarrow 1} (\frac{3}{1-x^3}-\frac{2}{1-x^2})
Final Answer
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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