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

Question

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

I am trying to evaluate this and so far this is what i have done -

$$ \lim _{x \rightarrow 1}\left(\frac{1}{1-x}-\frac{3}{1-x^{3}}\right) = \lim _{x\to 1}\left(\frac{\left({1-x}\right)\left(1+x^2+x-3\right)}{\left(1-x\right)\left(1-x^3\right)}\right) $$ From above, I cancel out (1-x) from denominator and numerator.

$$= \lim _{x\to 1}\left(\frac{\left(x-1\right)\left(x+2\right)}{\left(1-x\right)\left(1+x+x^2\right)}\right)$$

Now I am stuck how to proceed further as x $\rightarrow 1$ .

in the last line, factor out $-1$ from $(x-1)$ in the numerator and cancel it with the $(1-x)$ in the denominator. Also, you should have $(1+x+x^2)$ in the denominator in the last line. — C Squared, Apr 29 '21 at 16:34
The next step is just that, for $x\neq 1,$ $$\frac{x-1}{1-x}=-1.$$ — Thomas Andrews, Apr 29 '21 at 16:38

score 0 · Accepted Answer · answered Apr 29 '21 at 16:39

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The next step is just that, for $x\neq 1,$ $$\frac{x-1}{1-x}=-1.$$

answered Apr 29 '21 at 16:39

Thomas Andrews

177,126

score 0 · Answer 2 · answered May 11 '21 at 05:45

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Here's a trick you can use: substitute x as 1+h as lim h tends to zero and then solve algebra . In the end all the 'h' terms will be zeros and you will get answer as 1

answered May 11 '21 at 05:45

Calculus guy

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Welcome to MSE. This should be a comment, not an answer. – José Carlos Santos May 11 '21 at 05:54

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

2 Answers2