Simplify.

$\underset{x\to 1}{lim}\frac{\sqrt{x-1}}{{x}^{2}+x}$and$\underset{x\to 1}{lim}+\frac{\sqrt{x-1}}{{x}^{2}+x}$

$\underset{x\to 1}{lim}\frac{\sqrt{x-1}}{{x}^{2}+x}$and$\underset{x\to 1}{lim}+\frac{\sqrt{x-1}}{{x}^{2}+x}$

comAttitRize8
2022-07-27
Answered

Simplify.

$\underset{x\to 1}{lim}\frac{\sqrt{x-1}}{{x}^{2}+x}$and$\underset{x\to 1}{lim}+\frac{\sqrt{x-1}}{{x}^{2}+x}$

$\underset{x\to 1}{lim}\frac{\sqrt{x-1}}{{x}^{2}+x}$and$\underset{x\to 1}{lim}+\frac{\sqrt{x-1}}{{x}^{2}+x}$

You can still ask an expert for help

esbalatzaj

Answered 2022-07-28
Author has **15** answers

in order for to exist then both and $\underset{x\to 1}{lim}-\frac{\sqrt{x-1}}{{x}^{2}+x}$ must exist. Since $\frac{\sqrt{x-1}}{{x}^{2}+x}$ is undefined for x < 1 then doesnot exist. however does exist since it is a one sided limit, the limit is zero.

Almintas2l

Answered 2022-07-29
Author has **6** answers

$\underset{x\to 1}{lim}\ufeff\frac{\sqrt{x-1}}{{x}^{2}+x}=\frac{\sqrt{1-1}}{1+1}=0$

$\underset{x\to 1}{lim}-\frac{\sqrt{x-1}}{{x}^{2}+x}=\underset{x\to 1}{lim}+\frac{\sqrt{x-1}}{{x}^{2}+x}=0=\underset{x\to 1}{lim}\frac{\sqrt{x-1}}{{x}^{2}+x}$

$\underset{x\to 1}{lim}-\frac{\sqrt{x-1}}{{x}^{2}+x}=\underset{x\to 1}{lim}+\frac{\sqrt{x-1}}{{x}^{2}+x}=0=\underset{x\to 1}{lim}\frac{\sqrt{x-1}}{{x}^{2}+x}$

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The figure is something like:

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Finding a certain antiderivative

The problem says:

If $f(x)=\frac{{b}_{2}}{{x}^{2}}+\frac{{b}_{3}}{{x}^{3}}+...+\frac{{b}_{m}}{{x}^{m}}$ find a function g with ${g}^{\prime}=f$..

This problem is stated in the differentiation part of the book, integration comes later.

I tried starting with a simple example and trying to find a pattern from there:

If $f(x)=\frac{{b}_{2}}{{x}^{2}}$

Then we can try: $g(x)=-\frac{{b}_{2}}{x}$

${g}^{\prime}(x)=-{b}_{2}\left(\frac{-1}{{x}^{2}}\right)=\frac{{b}_{2}}{{x}^{2}}=f(x)$

Then if $f(x)=\frac{{b}_{2}}{{x}^{2}}+\frac{{b}_{3}}{{x}^{3}}$.

I tried: $g(x)=-\frac{{b}_{2}}{x}-\frac{{b}_{3}}{{x}^{3/2}}$

${g}^{\prime}(x)=-{b}_{2}\left(\frac{-1}{{x}^{2}}\right)-{b}_{3}\left(\frac{-\frac{3}{2}{x}^{1/2}}{{x}^{3}}\right)$

Which is not what I wanted. The denominator in the second summand is ok, but the numerator is not what I need.

But I'm stuck on what I need to look for, any hints for deriving the correct pattern would be appreciated.

The problem says:

If $f(x)=\frac{{b}_{2}}{{x}^{2}}+\frac{{b}_{3}}{{x}^{3}}+...+\frac{{b}_{m}}{{x}^{m}}$ find a function g with ${g}^{\prime}=f$..

This problem is stated in the differentiation part of the book, integration comes later.

I tried starting with a simple example and trying to find a pattern from there:

If $f(x)=\frac{{b}_{2}}{{x}^{2}}$

Then we can try: $g(x)=-\frac{{b}_{2}}{x}$

${g}^{\prime}(x)=-{b}_{2}\left(\frac{-1}{{x}^{2}}\right)=\frac{{b}_{2}}{{x}^{2}}=f(x)$

Then if $f(x)=\frac{{b}_{2}}{{x}^{2}}+\frac{{b}_{3}}{{x}^{3}}$.

I tried: $g(x)=-\frac{{b}_{2}}{x}-\frac{{b}_{3}}{{x}^{3/2}}$

${g}^{\prime}(x)=-{b}_{2}\left(\frac{-1}{{x}^{2}}\right)-{b}_{3}\left(\frac{-\frac{3}{2}{x}^{1/2}}{{x}^{3}}\right)$

Which is not what I wanted. The denominator in the second summand is ok, but the numerator is not what I need.

But I'm stuck on what I need to look for, any hints for deriving the correct pattern would be appreciated.

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