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Evaluate the question of the following limits.

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

reshown
0 after applying L hopital rule

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

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

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

$=∞$

Note :

by

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.

@ankitgupta.1729  thanks sir for confirming .

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.
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 $\large +\infty$.

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