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Evaluate: $$ \lim_{x\to0}\frac{\sqrt[5]{2x^2 + 10x + 1} - \sqrt[7]{x^2 + 10x + 1}}{x} $$

I want to find the limit above. So far it seems a good idea to rationalize the nominator in hope for $x$ to appear in the nominator so that it cancels out $x$ in the denominator.

I've been able to show that the limit is equal to $4\over 7$ by using L'Hospial's rule: $$ \lim_{x\to 0}f(x) = {1\over 5}\left(x^2 + 10 x + 1\right)^{-{4\over 5}}(4x + 10) - {1\over 7}\left(x^2 + 10x + 1\right)^{-{6\over 7}}(2x + 10) = {10\over 5} - {10\over 7} = {4\over 7} $$

How to find that limit without using derivatives, but rather rationalizing the nominator (or potentially using another approach, but also with no derivatived involved)?

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    $\begingroup$ Use Maclaurin series: $\sqrt[5]{2x^2+10x+1}=1+2x+\cdots$ etc. $\endgroup$ – Angina Seng Apr 9 '19 at 16:08
  • $\begingroup$ Whether the first-order Maclaurin series "use derivatives" may be debatable. $\endgroup$ – Robert Israel Apr 9 '19 at 16:18
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Because $$\frac{\sqrt[5]{2x^2 + 10x + 1} - \sqrt[7]{x^2 + 10x + 1}}{x}=\frac{\sqrt[5]{2x^2 + 10x + 1}-1 - \left(\sqrt[7]{x^2 + 10x + 1}-1\right)}{x}=$$ $$=\tfrac{2x+10}{\left(\sqrt[5]{2x^2 + 10x + 1}\right)^4+\left(\sqrt[5]{2x^2 + 10x + 1}\right)^3+\left(\sqrt[5]{2x^2 + 10x + 1}\right)^2+\left(\sqrt[5]{2x^2 + 10x + 1}\right)+1}-$$ $$-\tfrac{x+10}{\sqrt[7]{(x^2 + 10x + 1)^6}+\sqrt[7]{(x^2 + 10x + 1)^5}+\sqrt[7]{(x^2 + 10x + 1)^4}+\sqrt[7]{(x^2 + 10x + 1)^3}+\sqrt[7]{(x^2 + 10x + 1)^2}+\sqrt[7]{x^2 + 10x + 1}+1}\rightarrow$$ $$\rightarrow\frac{10}{5}-\frac{10}{7}=\frac{4}{7}.$$

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    $\begingroup$ +1, definitely easier than mine. $\endgroup$ – farruhota Apr 9 '19 at 17:38
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You find separately $$ a=\lim_{x\to0}\frac{\sqrt[5]{2x^2 + 10x + 1}-1}{x} $$ and $$ b=\lim_{x\to0}\frac{\sqrt[7]{x^2 + 10x + 1}-1}{x} $$ If they both exist finite, then your limit is $a-b$.

The first one can be easily rationalized; let me write $f(x)=\sqrt[5]{2x^2 + 10x + 1}$ for simplicity: $$ a=\lim_{x\to0}\frac{2x^2+10x+1-1}{x}\frac{1}{f(x)^4+f(x)^3+f(x)^2+f(x)+1}=10\frac{1}{5}=2 $$

Similarly, the second limit is $10/7$, so the sought limit is $$ a-b=2-\frac{10}{7}=\frac{4}{7} $$

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  • $\begingroup$ Dear Teacher, I want to ask You a favor/help.. I'm aware that this place is not appropriate. Is there an error in my question/solution (in its most recent form)? If there are errors, can you help to correct the error? Thank you very much...math.stackexchange.com/q/3185610/460967 $\endgroup$ – lone student Apr 13 '19 at 19:57
  • $\begingroup$ @Student I find that way is quite alike to torture, which is prohibited by several international conventions. The photo on the blackboard is like a bad nightmare. Why avoiding derivatives that make the computation much easier? $\endgroup$ – egreg Apr 13 '19 at 20:25
  • $\begingroup$ I'm sorry. I think, I didn't understand your comment correctly...:) (My english is insufficient) (The first sentence Your comment, Does it apply to my solution? Or for the blackboard? $\endgroup$ – lone student Apr 13 '19 at 21:30
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    $\begingroup$ @Student Computing that limit the hard way (without derivatives) is just painful. And that photo of a blackboard shows a very cryptic way to do the job. $\endgroup$ – egreg Apr 13 '19 at 21:36
  • $\begingroup$ I understand, I understood.. I just want to know the truth. I mean, the method I used was prohibited by several international conventions. Is it correct? As I understand it, my method is not rigorous.. $\endgroup$ – lone student Apr 13 '19 at 21:47
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Let $a=2x^2 + 10x + 1; b=x^2 + 10x + 1$.

Then: $$\\ \lim_{x\to0}\frac{\sqrt[5]{2x^2 + 10x + 1} - \sqrt[7]{x^2 + 10x + 1}}{x}=\\ \lim_{x\to0}\frac{a^{1/5} - b^{1/7}}{x}=\\ \lim_{x\to0}\frac{a^{1/5} - b^{1/5}+b^{1/5}-b^{1/7}}{x}=\\ \lim_{x\to0}\frac{a^{1/5} - b^{1/5}}{x}+\lim_{x\to0} \frac{b^{1/5}-b^{1/7}}{x}=A+B.$$ Hence: $$A=\lim_{x\to0}\frac{(a^{1/5} - b^{1/5})(a^{4/5}+a^{3/5}b^{1/5}+a^{2/5}b^{2/5}+a^{1/5}b^{3/5}+b^{4/5})}{x(a^{4/5}+a^{3/5}b^{1/5}+a^{2/5}b^{2/5}+a^{1/5}b^{3/5}+b^{4/5})}=\\ \lim_{x\to0}\frac{a - b}{x(a^{4/5}+a^{3/5}b^{1/5}+a^{2/5}b^{2/5}+a^{1/5}b^{3/5}+b^{4/5})}=\\ \lim_{x\to0}\frac{x}{a^{4/5}+a^{3/5}b^{1/5}+a^{2/5}b^{2/5}+a^{1/5}b^{3/5}+b^{4/5}}=\frac05=0.$$ Similarly: $$B=\lim_{x\to0} \frac{b^{1/5}-b^{1/7}}{x}=\lim_{x\to0} \frac{b^{1/7}(b^{2/35}-1)}{x}=\lim_{x\to0} \frac{b^{2/35}-1}{x}=\\ \lim_{x\to0} \frac{((b^2)^{1/35}-1)[(b^2)^{34/35}+(b^2)^{33/35}+\cdots +(b^2)^{1/35}+1]}{x[(b^2)^{34/35}+(b^2)^{33/35}+\cdots +(b^2)^{1/35}+1]}=\\ \lim_{x\to0} \frac{b^2-1}{x[(b^2)^{34/35}+(b^2)^{33/35}+\cdots +(b^2)^{1/35}+1]}=\\ \lim_{x\to0} \frac{x^3+20x^2+102x+20}{(b^2)^{34/35}+(b^2)^{33/35}+\cdots +(b^2)^{1/35}+1}=\\ \frac{20}{35}=\frac47.$$

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Standard but not so popular limit formula $$\lim_{x\to a} \frac{x^n-a^n} {x-a} =na^{n-1}$$ is of great help here.

Let $$u=1+10x+2x^2,v=1+10x+x^2$$ so that both $u, v$ tend to $1$ as $x\to 0$. Moreover note that $(u-1)/x,(v-1)/x$ both tend to $10$. The limit in question can be written as $$\lim_{u\to 1}\frac{u^{1/5}-1^{1/5}}{u-1}\cdot\frac{u-1}{x}-\frac{v^{1/7}-1^{1/7}}{v-1}\cdot\frac{v-1}{x}$$ which by the standard limit formula equals $$\frac{1}{5}\cdot 10-\frac{1}{7}\cdot 10=\frac{4}{7}$$

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