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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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