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I have a problem with calculation of the limit:

$$\lim\limits_{x \to 7} \frac{\sqrt{x+2} - \sqrt[3]{x+20}}{\sqrt[4]{x+9} - 2}$$

Is there a way to calculate it? How can I do it?

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You can calculate separately the limits $$\frac{\sqrt{x+2}-3}{x-7}$$ $$\frac{\sqrt[3]{x+20}-3}{x-7}$$ $$\frac{\sqrt[4]{x+9}-2}{x-7}$$ For this you use the trick of multiplying by conjugates I dont know if you know it. For the middle one use $$(x-y)(x^2+xy+y^2)=x^3-y^3$$ and for the first and last, $$(x-y)(x+y)=x^2-y^2$$ once for the first and use it two times successively for the last.

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With Taylor's formula at order $1$:

Set $x=7+h$. You obtain \begin{align} \frac{\sqrt{x+2} - \sqrt[3]{x+20}}{\sqrt[4]{x+9} - 2}&=\frac{\sqrt{9+h} - \sqrt[3]{27+h}}{\sqrt[4]{16+h} - 2}=\frac{3\sqrt{1+\frac h9} - 3\sqrt[3]{1+\frac h{27}}}{2\sqrt[4]{1+\frac h{16}} - 2}\\ &=\frac{\bigl(3+\frac h6+o(h)\bigr)-\bigl(3+\frac h{27}+o(h)\bigr)}{2+\frac h{32}+o(h)-2}=\frac{\frac{7h}{54}+o(h)}{\frac h{32}+o(h)}=\frac{\frac{7}{54}+o(1)}{\frac 1{32}+o(1)}\to \frac{112}{27}. \end{align}

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  • $\begingroup$ Essentially this has something to do with derivative, but OP is not allowed to use it. $\endgroup$ – user284331 Nov 29 '17 at 17:52
  • $\begingroup$ Yes, it's right... I have to find the way to do it without derivatives. $\endgroup$ – Nikolai Paukov Nov 29 '17 at 18:00
  • $\begingroup$ I think what the OP is looking for is a direct epsilon-delta calculation of the limit from the definition. $\endgroup$ – Mathemagician1234 Nov 29 '17 at 18:07
  • $\begingroup$ Mathemagician1234, Yes, I am! Calculation with derivatives is not very diffucult, but I can't use them. $\endgroup$ – Nikolai Paukov Nov 29 '17 at 18:09
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Hint:

If $\sqrt{x+2}=a,\sqrt[3]{x+20}=b,\sqrt[4]{x+9}=c$

LCM$(2,3)=6$

$a^6-b^6=(a-b)(\cdots)$

Similarly $c^4-2^4=(c-2)\cdots$

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This is a long comment.

The approximation $(1+y)^{1/n}\approx 1+\frac{y}{n}$ used in Bernard's answer doesn't require calculus for natural $n$, because the binomial theorem gives $(1+\frac{y}{n})^n\approx 1+y$. One need only expand the product of $n$ terms with algebra.

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  • $\begingroup$ A rigourous approximation formula requires a complementary $o(h)$ or $O(h^2)$ term, and of course Taylor's formula makes it easier to justify. $\endgroup$ – Bernard Nov 29 '17 at 18:40
  • $\begingroup$ @Bernard Yes, but we can get the $y^2$ term combinatorially as well. $\endgroup$ – J.G. Nov 29 '17 at 18:40
  • $\begingroup$ I agree, but is it worth the effort? $\endgroup$ – Bernard Nov 29 '17 at 18:41
  • $\begingroup$ @Bernard Only if someone says, "I agree $f(7+\epsilon)\approx L$, but how does that prove $f\to L$?" $\endgroup$ – J.G. Nov 29 '17 at 18:44
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Let $f(x)=\sqrt{x+2}-\sqrt[3]{x+20}$ and $g(x)=\sqrt[4]{x+9}-2$, then \begin{align*} \lim_{x\rightarrow 7}\dfrac{\sqrt{x+2}-\sqrt[3]{x+20}}{\sqrt[4]{x+9}-2}&=\lim_{x\rightarrow 7}\dfrac{f(x)-f(7)}{x-7}\lim_{x\rightarrow 7}\dfrac{x-7}{g(x)-g(7)}\\ &=f'(7)\times\dfrac{1}{g'(7)}. \end{align*}

Alternatively, if we let \begin{align*} \dfrac{\sqrt{x+2}-\sqrt[3]{x+20}}{\sqrt[4]{x+9}-2}&=\dfrac{\dfrac{u^{1/2}-v^{1/2}}{u^{1/3}+u^{1/6}v^{1/6}+v^{1/3}}}{\dfrac{x-7}{(x+9)^{1/4}+2}\dfrac{1}{(x+9)^{1/2}+4}}, \end{align*} where $u=(x+2)^{3}$, $v=(x+20)^{2}$, now $u^{1/2}-v^{1/2}=\dfrac{u-v}{u^{1/2}+v^{1/2}}$, where $u-v=(x-7)(x^{2}+12x+56)$.

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  • $\begingroup$ This is not L'Hospital, but the very definition of derivative. $\endgroup$ – user284331 Nov 29 '17 at 17:34
  • $\begingroup$ Unfortunately, I can't use derivatives. $\endgroup$ – Nikolai Paukov Nov 29 '17 at 17:49
  • $\begingroup$ I think right now it works, but messy. $\endgroup$ – user284331 Nov 29 '17 at 18:27
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Using $x^n - y^n = (x - y)\sum_{k=0}^n x^ky^{n-k}$, we have $$ \sqrt{x+2}-\sqrt[3]{x+20} = \frac{(x+2)^3-(x+20)^2}{\sum_{k=0}^6(x+2)^{k/2}(x+20)^{2-k/3}} \\ \sqrt[4]{x+9}-2 = \frac{x-7}{\sum_{k=0}^4(x+9)^{k/4}2^{4-k}} $$ Notice that $(x+2)^3-(x+20)^2 = (x - 7)(x^2+12x+56)$, so $$ \begin{aligned} \frac{\sqrt{x+2}-\sqrt[3]{x+20}}{\sqrt[4]{x+9}-2} & = \frac{(x - 7)(x^2+12x+56)/\sum_{k=0}^6(x+2)^{k/2}(x+20)^{2-k/3}}{(x-7)/\sum_{k=0}^4(x+9)^{k/4}2^{4-k}} \\ & = (x^2+12x+56)\frac{\sum_{k=0}^4(x+9)^{k/4}2^{4-k}}{\sum_{k=0}^6(x+2)^{k/2}(x+20)^{2-k/3}} \end{aligned} $$ so $$ \begin{aligned} \lim_{x\to 7}\frac{\sqrt{x+2}-\sqrt[3]{x+20}}{\sqrt[4]{x+9}-2} & = (7^2+12\cdot7+56)\frac{\sum_{k=0}^416^{k/4}2^{4-k}}{\sum_{k=0}^69^{k/2}27^{2-k/3}} \\ & = 189\cdot\frac{2^4}{27^2} = \frac{112}{27} \end{aligned} $$

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$$\lim\limits_{x \to 7} \frac{\sqrt{x+2} - \sqrt[3]{x+20}}{\sqrt[4]{x+9} - 2}=\lim\limits_{x \to 7} \frac{\sqrt{9+x-7} - \sqrt[3]{27+x-7}}{\sqrt[4]{16+x-7} - 2}=$$ $$=\lim\limits_{x \to 7} \frac{3\sqrt{1+\frac{x-7}{9}} - 3\sqrt[3]{1+\frac{x-7}{27}}}{2\sqrt[4]{1+\frac{x-7}{16}} - 2}=\frac{3}{2}\lim\limits_{x \to 7}\frac{\left(1+\frac{x-7}{9}\right)^{\frac{1}{2}}-\left(1+\frac{x-7}{27}\right)^{\frac{1}{3}}}{\left(1+\frac{x-7}{16}\right)^\frac{1}{4}-1}$$

Now recall that $\lim\limits_{t \to t_0}\left[1+\left(f(t)-f(t_0)\right)\right]^{\alpha}=1+\alpha\left[f(t)-f(t_0)\right]+o\left(f(t)-f(t_0)\right)$ if $f(t)\to f(t_0)$ for $t \to t_0$

$$\frac{3}{2}\lim\limits_{x \to 7}\frac{\left(1+\frac{x-7}{9}\right)^{\frac{1}{2}}-\left(1+\frac{x-7}{27}\right)^{\frac{1}{3}}}{\left(1+\frac{x-7}{16}\right)^\frac{1}{4}-1}=\frac{3}{2}\lim\limits_{x \to 7}\frac{1+\frac{1}{2}\frac{x-7}{9}-1-\frac{1}{3}\frac{x-7}{27}+o\left(x-7\right)}{1+\frac{1}{4}\frac{x-7}{16}-1+o\left(x-7\right)}=$$

$$=\frac{3}{2}\lim\limits_{x \to 7}\frac{\frac{x-7}{18}-\frac{x-7}{81}+o\left(x-7\right)}{\frac{x-7}{64}+o\left(x-7\right)}=96\lim\limits_{x \to 7}\frac{(x-7)\left(\frac{1}{18}-\frac{1}{81}\right)+o\left(x-7\right)}{(x-7)\left[1+o\left(x-7\right)\right]}=96\cdot \frac{28}{648}=\frac{112}{27}$$.

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The unpopular limit formula $$\lim_{x\to a} \frac{x^{n} - a^{n}} {x-a} =na^{n-1}\tag{1}$$ is your friend here. First divide the numerator and denominator of the expression by $x-7$ and note that the denominator itself becomes a fraction $$\frac{\sqrt[4]{x+9}-2} {x-7}=\frac{t^{1/4}-16^{1/4}}{t-16}$$ where $t=x+9\to 16$. By the limit formula $(1)$ the above fraction tends to $(1/4)16^{-3/4}=1/32$.

Next consider the numerator which has now become $$\frac{\sqrt{x+2}-\sqrt[3]{x+20}}{x-7}=\frac{u^{1/2}-9^{1/2}}{u-9}-\frac{v^{1/3}-27^{1/3}}{v-27}$$ where $u=x+2\to 9,v=x+20\to 27$. By the limit formula $(1)$ the above expression tends to $(1/2)9^{-1/2}-(1/3)27^{-2/3}=7/54$. The final answer is now $(7/54)/(1/32)=112/27$.

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