Applied Mathematics Calculus and Limit Question

Question

Evaluate the question of the following limits. 

 

• $\lim_{x\rightarrow 1} \frac{x}{(x-1)^{2}}$

Comments

0 after applying L hopital rule

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@jugnu1337 how did  L’hopital rule applicable here ? is it a $\frac{0}{0}$ or $\frac{\infty}{\infty}$ form?

Answer 1

$\lim_{x\rightarrow 1}$   $x*\frac{1}{(x-1)^2}$

$\lim_{x\rightarrow 1}x$   $*\lim_{x\rightarrow 1} \frac{1}{(x-1)^2}$

$=∞$

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Note :

Comments

@ANSR1010 it should be positive infinite. as there is two possibilities here if for left hand limit and right hand limit the value of infinite comes $-\infty$ and $+\infty$ then left hand limit and right hand limit are not equal so by defination limit does not exists. for example , lim x->0 1/x , here limit does not exists due to left hand limit and right hand limit value are not equal .

For the given problem in the first portion  lim x->1 1/(x-1) lim does not exist for this case. so your solution is not correct for that expression.

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A good read

https://math.stackexchange.com/a/127671/913269

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$\large \lim_{x->1}\frac{1}{x-1}$ how for this limit exists?

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Bro, you are unnecessary complicating it. Look again at the example you gave. My answer is same.

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bro but how limit exists for the expression  $\large \lim_{x->1}\frac{1}{x-1}$ ? For this expression limit does not exist.

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what’s the denominator in question? Look again it's $(x-1)^2$ and if has been $(x-1)$ then all your saying is correct.

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Bro what I am saying is you divide the problem in two part 

1. $\large \lim_{x->1}\frac{1}{x-1}$ say it is $L1$
2. $\large \lim_{x->1}\frac{1}{(x-1)^{2}}$ say It is $L2$

Now you adding the result of these two to get the answer of the desired result. That is fine.

Now for $L2$ the answer comes as $+\infty$ ok that is fine .

Now for L1 limit does not exists.

Now my doubt is it is possible to add these two result as one of the result is limit does exists ?

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Is this ok for you ?

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@ASNR1010 now it is fine.

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Bro I don’t know why you are taking cases in your answer and ∞ and +∞ both are same their is no difference as you said earlier take +. It’s limit always exist. Their is no need of checking it. By seeing the question only you can say that.

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No bro they are not same . you can think like that for the example $\large \lim_{x->0}\frac{1}{x}$ , why for this case limit does not exist ? It is because of the reason that Left hand limit and right hand limit value comes as $\large +\infty$ and $\large -\infty$ .

https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2006/acebd5cc8fe0315270d486685739d08f_lec2.pdf

read this bro . This result help to proof many infinite discontinuity. 

we have to check for  $+\infty$ and $\large -\infty$ because as the limit value comes here as infinity so if right hand limit and left hand limit comes as  $+\infty$ and $\large -\infty$ then the limit does not exists so we have to careful to check infinite discontinuity.

we can think of  $+\infty$ and $\large -\infty$ as $\large \infty$ when the context of sign does not change the result but here depending on the sign results are affecting so we have to cautious about this.

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Don’t want to comment. My above comments are fine for justifying what I’m saying. tag anyone for clarification.

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I have already justified my point If you feel anything is wrong then please mention anyone to justify.

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and well known site for mathematics is wrong because they miss +. right?

and more funnier part is I am not pro member of this site but you can see the solution hidden. It is exactly same as my answer. Look below screenshot.

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@ASNR1010 i think you are either don’t understand what i was explained or you don’t want to understand .

Infinity is a general term generally used to represent a very large boundless number (may not be a proper definition) .

Now as it is also number we can have both positive and negative infinity depending on in which side of $0$ it grows .

Now It is not wrong to say only $\infty$ if the context is clear . Here i advised you to write  $+\infty$ as in the context it is more suitable also from the “well known site of mathematics” the pic you provided if you look carefully it is showing the value increases towards positive infinity .

And the solution you have provided earlier there i had one doubt which asked you that can we add two limits where one limits evaluated to limit does not exists and another $+\infty$ to which you changed  the approach which looks fine to me . 

I also explained earlier why $+\infty$ and $-\infty$ is important in the context of limit as it will help us to evaluate the limit correctly if exists or identify if doesn’t.

Hope you will try to think what i am explained and correct me if there is anything wrong or illogical.

 

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Writing $\lim_{x \rightarrow a} (f(x)+g(x)) = \lim_{x \rightarrow a} f(x) + \lim_{x \rightarrow a} g(x)$

is valid only if both $\lim_{x \rightarrow a} f(x)$ and $\lim_{x \rightarrow a} g(x)$ exist

Or we can say ,if $\lim_{x \rightarrow a} f(x) = P$ and $\lim_{x \rightarrow a} g(x) = Q$

then we can write $\lim_{x \rightarrow a} (f(x)+g(x)) = P + Q$

It can be proved also.

So, here,

Writing $\lim_{x \rightarrow 1} \frac{x}{(x-1)^2} =\lim_{x \rightarrow 1} \left(\frac{1}{(x-1)} + \frac{1}{(x-1)^2} \right) =  \lim_{x \rightarrow 1}\frac{1}{(x-1)}  + \lim_{x \rightarrow 1} \frac{1}{(x-1)^2}$

is incorrect because $ \lim_{x \rightarrow 1}\frac{1}{(x-1)}$ does not exist.

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@ankitgupta.1729  thanks sir for confirming .

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Infinity has different meaning in different context. For example, in real analysis, infinity is considered as a “symbol” means  $+\infty$ is a symbol and $-\infty$ is a symbol and there is no notion of equality between symbols, on the other side, at some places, infinity is considered as a number. In set theory, we study about infinite sets and here there is no notion of positive infinite sets or negative infinite sets. And there might be other places where infinity has different meaning. So, to understand whether $+ \infty$ is same as $-\infty,$ we have to understand the context, we have to understand the meaning of infinity.

In probability theory, some people say probability is always positive but it is not true. Probability can be negative. I can show it in a very good book where notion of negative probability is given. We say positive probability because we follow the theory which is based on the assumption that probability has a value between 0 and 1 (including both). If we break this assumption then probability can be negative too and it might be possible on some other planet, there is a notion of negative probability.

Answer 2

First for this case we have to check if limit exist or not . So we have to calculate both left hand limit and right hand limit and to exist the limit they has to be same.

LHL:-

$\large \lim_{x->1^{-}}\frac{x}{(x-1)^{2}}$

$\large \lim_{x->1^{-}}$ we are coming toward 1 from from fraction side means say value of x is 0.000000099 like that.

So from the limit we can observe numerator is positive and in denominator (x-1) is negative as value of x is close to 1 from left side . so $\large (x-1)^{2}$ is positive but very small.

so, as x tends to $\large 1^{-}$,then  $\large (x-1)^{2}->0^{+}$

So, $\large \lim_{x->1^{-}}\frac{x}{(x-1)^{2}}$ = $\large +\infty$

RHL:-

$\large \lim_{x->1^{+}}\frac{x}{(x-1)^{2}}$

$\large \lim_{x->1^{+}}$ we are coming toward 1 from from right side means say value of x is 1.00000001 like that.

So from the limit we can observe numerator is positive and in denominator (x-1) is positive as value of x is close to 1 from right side . so $\large (x-1)^{2}$ is positive but very small.

so, as x tends to $\large 1^{+}$ ,then $\large (x-1)^{2}->0^{+}$.

$\large \lim_{x->1^{+}}\frac{x}{(x-1)^{2}}$=$\large +\infty$

So,

both left hand limit and right hand limit exists and both are equal to $\large +\infty$ .

So limit exist and the value of the limit is $\large +\infty$.