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Struggled with this question, looked over examples still couldn't figure it out.


Question: Find the limit
$\lim\limits_{x \to 1} \frac{2-x}{(x-1)^2}$

What i got to: $\lim\limits_{x \to 1} {2-x} \frac {1}{(x-1)(x+1)}$

Understanding of infinite limits is "limited." For this question, am i supposed to do the right side first of the limit and then left side and then see if the outputs match, if so the limit exists?

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    $\begingroup$ What is the issue here ? You get $\frac 10$ so limit does not exists (it is infinity with undefined sign). What is bothering you ? Also $(x-1)^2=(x-1)(x-1)\neq x^2-1$ $\endgroup$ – zwim May 30 '20 at 2:58
  • $\begingroup$ The "naive" way is $\frac {2-x}{(x-1)^2} = \frac 10$. As $(x-1)^2 > 0$ we have that the answer should be $\lim \frac {2-x}{(x-1)^2} =+\infty$ so we need to set up our delta $N$ prove that for any $N$ there is a $\delta> 0$ so that $0< |x-1| < \delta$ then $\frac {2-x}{(x-1)^2} > N$ $\endgroup$ – fleablood May 30 '20 at 3:37
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When we put $x=1$ in the limit, we get $$\frac{1}{0}$$ This is not an indeterminate form. Therefore we consider the right-hand and left-hand limits respectively. For the right-hand limit, the limit approaches positive infinity as both the top and bottom of the fraction are positive. For the left-hand limit, the limit also approaches positive infinity because both the top and bottom of the fraction are positive due to the square on the bottom. Therefore we conclude $\frac{2-x}{(x-1)^2}\to+\infty$ as $x\to 1$

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A limit

$$\lim_{x\to a} f(x)$$

exists if and only if it is equal to a number. Note that $\infty$ is not a number.

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The function is not defined for $x = 1$. In addition, no sign changes around $x = +1$ and $x = -1$ so the denominator grows quickly. So when $x \rightarrow 1$ the limit does not exist.

So we can calculate separately

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

And the denominator grows without quota . You can use the same reasoning for growth from the right.

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

In the right part of the multiplication we get $2-1 = 1$. And on the othe for x approaching 1 from the right, denominator is a positive value that approaches 0 and $\lim _{x\to \:1+}\left(\frac{1}{\left(x-1\right)^2}\right) =\infty \:$

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Let's find both the Right-hand limit and Left-hand limit separately: First, the left hand limit: $$\lim_{x\to 1^-} \frac{2-x}{(x-1)^2}$$ $$=\lim_{h\to 0^+}\frac{2-(1-h)}{[(1-h)-1]^2}$$ $$=\lim_{h\to 0^+}\frac{1-h}{h^2}$$ $$\implies \text{Left-hand limit}\to +\infty$$ Now the right hand limit: $$\lim_{x\to 1^+}\frac{2-x}{(x-1)^2}$$ $$=\lim_{h\to 0^+}\frac{2-(1+h)}{(1+h-1)^2}$$ $$=\lim_{h\to 0^+}\frac{1-h}{h^2}$$ $$\implies \text{Right-hand limit} \to +\infty$$ Since both the limits are non existent, and tend to $+\infty$ you can conclude that the limit does not exist at $x=1$.


P.S. : You have wrongly written $(x-1)^2= (x-1)(x+1)$ ( as pointed out by @zwim)

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The "naive" way is $\frac {2-x}{(x-1)^2} = \frac 10$. As $(x-1)^2 > 0$ we have that the answer should be $\lim \frac {2-x}{(x-1)^2} =+\infty$ so we need to set up our delta $N$ prove that for any $N$ there is a $\delta> 0$ so that $0< |x-1| < \delta$ then $\frac {2-x}{(x-1)^2} > N$

So we want $0 < x-1< \delta \implies$

$1 -\delta < x < 1+ \delta$ and if we assume $\delta < 1$ then that implies

$1 -\delta < 2-x < 1+\delta$ and $0 < (x-1)^2 < \delta^2$ so

$\frac {2-x}{(x-1)^2} > \frac {2-x}{\delta^2} >\frac {1-\delta}{\delta^2} > \frac {1-\delta}{\delta} =\frac 1\delta - 1$(remember we are assuming $0 < \delta < 1$)

So so if we choose $\delta$ so that $\frac 1\delta - 1 > N$ or in other words if $\delta < \frac 1{N+1}$ we are done. Such a delta so that $\delta < \frac 1{N+1}$ and $0 < \delta < 1$ can always be found. (Well, it can if $N > -1$ but as we are trying to prove this for large $N$ we substitute $N' = \max(N,0)$ and do it.)

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