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I need to find the following limit: $$\lim_{x\to -1}\frac{x^2 - 2x - 3 }{x + 1}$$

The polynomial is simplified to $\dfrac{(x+1)(x-3)}{x+1}$

Hello, I can solve this by plugging in the value $-1$ after factoring the polynomial and simplifying (which then equals $-4$) but I'm not sure how to solve it using left and right handed limits and using piece wise defined functions.

The problem asks to "Consider left and right sided limits and use piecewise defined functions."

Any help would be appreciated. Thanks!

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    $\begingroup$ There is no need to consider left and right sided limits, as your function coincides with $x\mapsto x-3$ for every $x\ne-1$. Your solution is correct. $\endgroup$
    – egreg
    Commented Jun 26, 2014 at 20:32
  • $\begingroup$ Okay, thank you. What about lim x-> 3 which is the next question. Would I need to consider right,left, piecewise for that? Same f(x) is used. I'm not sure why my teacher would say "consider" if it isn't even required. $\endgroup$
    – androidguy
    Commented Jun 26, 2014 at 20:37
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    $\begingroup$ For $x \to 3$ you have a continuous function, so can just evaluate it and get $0$ $\endgroup$ Commented Jun 26, 2014 at 20:38
  • $\begingroup$ Thanks for the help! When do you use piecewise functions to solve limits? $\endgroup$
    – androidguy
    Commented Jun 26, 2014 at 20:40

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Your analysis is correct: for $x\ne-1$, you have $$ f(x)=\frac{x^2 - 2x - 3 }{x + 1}=\frac{(x+1)(x-3)}{x+1}=x-3 $$ Since the function $g(x)=x-3$ is everywhere continuous, its limit for $x\to-1$ is $g(-1)=-4$, which means also that $$ \lim_{x\to-1}f(x)=\lim_{x\to-1}g(x)=-4. $$

There is no problem for $\lim_{x\to3}f(x)$, since the function is continuous at $3$, so $\lim_{x\to3}f(x)=f(3)=0$.

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  • $\begingroup$ Great answer, why is it that the hint on this problem would be "Consider right and left sided and piecewise" When would these be used? Is there anyway to incorporate these in my answer? $\endgroup$
    – androidguy
    Commented Jun 26, 2014 at 20:57
  • $\begingroup$ @androidguy I really don't know. Maybe trying to deceive students by making them believe that math is difficult? $\endgroup$
    – egreg
    Commented Jun 26, 2014 at 20:59
  • $\begingroup$ Haha, that's funny. It's difficult for me so it just made it more confusing. Thanks for the help. $\endgroup$
    – androidguy
    Commented Jun 26, 2014 at 21:01

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