# Finding the limit question

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?

• 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 $(x-1)^2=(x-1)(x-1)\neq x^2-1$ – zwim May 30 '20 at 2:58
• 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 $\frac {2-x}{(x-1)^2} > N$ – fleablood May 30 '20 at 3:37

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 $$x\to 1$$

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.

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

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)

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.)