# Computing : $\lim_{x\to 1} \frac{\sqrt{4x^2+5}-3}{\sqrt[3]{x^4}-1}$ without L'Hopital or derivative [closed]

Can you please help me with this limit? I can´t use L'Hopital rule nor any derivatives. $$\lim_{x\to 1} \frac{\sqrt{4x^2+5}-3}{\sqrt[3]{x^2}-1}$$

Thank you and have a nice day!

PS: I´m really sorry that in my previous post I had some mistakes.

• The methods used in the answers/comments to your previous questions also apply here. Commented Jan 19, 2018 at 17:51
• Commented Jan 19, 2018 at 17:52
• math.stackexchange.com/questions/2612416/… Commented Jan 19, 2018 at 17:55
• Yes I can do it Commented Jan 19, 2018 at 18:20
• @labbhattacharjee what is the point of your link? Commented Jan 19, 2018 at 18:27

we have that $$\sqrt{4x^2+5}-3 =\frac{(\sqrt{4x^2+5}-3)(\sqrt{4x^2+5}+3)}{\sqrt{4x^2+5}+3}=\frac{4(x^2-1)}{\left(\sqrt{4x^2+5}+3\right)}$$

Inserting $a=x^{2/3}$ is the geometric formula $$\frac{a^3-1}{a-1} = a^2+a+1\implies \frac{x^2-1}{\sqrt[3]{x^2}-1} = x^{4/3}+ x^{2/3}+1$$

Hence,

$$\color{blue}{ \frac{\sqrt{4x^2+5}-3}{\sqrt[3]{x^2}-1} = \frac{4(x^{4/3}+ x^{2/3}+1)}{\sqrt{4x^2+5}+3} \to 2~~as~~x\to 1}$$

• But OP stated: "I can´t use L'Hopital rule nor any derivatives." Commented Jan 19, 2018 at 18:01
• Thank you for the answer but I can´t use derivatives Commented Jan 19, 2018 at 18:01
• @StackTD see the edit now Commented Jan 19, 2018 at 18:15
• @GuyFsone That was what I was looking for, thank you and sorry for the mess Commented Jan 19, 2018 at 18:29
• @Impropio this is even much more simpler Commented Jan 19, 2018 at 18:33

$$\lim_{x\to 1} \frac{\sqrt{4x^2+5}-3}{\sqrt[3]{x^2}-1}$$

Multiply with the conjugate of the numerator and factor: $$\frac{\left(\sqrt{4x^2+5}-3\right)\color{blue}{\left(\sqrt{4x^2+5}+3\right)}}{\left(\sqrt[3]{x^2}-1\right)\color{blue}{\left(\sqrt{4x^2+5}+3\right)}} = \frac{4\left(x-1\right)\left(x+1\right)}{\left(\sqrt[3]{x}^2-1\right)\left(\sqrt{4x^2+5}+3\right)}$$ Hint: $\sqrt[3]{x}^2-1=\left(\sqrt[3]{x}-1\right)\left(\sqrt[3]{x}+1\right)$ and $\left(x-1\right)=\left(\sqrt[3]{x}-1\right)\left( \color{red}{\ldots} \right)$ via $a^3-b^3=\ldots$.

Can you complete the red dots? You can then simplify by cancelling the common factor $\sqrt[3]{x}-1$.

• Quick question. The part in blue shouldn't be + instead of -? Commented Jan 19, 2018 at 17:59
• @Impropio Yes of course, copy/paste-error. Thanks for noticing, I edited. Commented Jan 19, 2018 at 18:00
• Thank you for helping me! Commented Jan 19, 2018 at 18:30

Let $x^2=y+1$ with $y\to0$ and using that

$$y\to0\quad \implies(1+y)^a=1+ay+o(y)$$

we obtain

$$\frac{\sqrt{4x^2+5}-3}{\sqrt[3]{x^2}-1}=\frac{\sqrt{4y+9}-3}{\sqrt[3]{y+1}-1}=\frac{3\sqrt{\frac49y+1}-3}{\sqrt[3]{y+1}-1}=\frac{3\left(1+\frac4{18}y+o(y)\right)-3}{1+\frac13y+o(y)-1}=\frac{3+\frac23y+o(y)-3}{\frac13y+o(y)}=\frac{\frac23y+o(y)}{\frac13y+o(y)}\to\frac23 \cdot3=2$$

• Thank you for the answer but I've not seen the part of (1+y)^a in this course Commented Jan 19, 2018 at 18:00
• Then the others solutions by radicals are good for you!
– user
Commented Jan 19, 2018 at 18:05

UPDATE: Apologies, I misread your initial condition stating that you cannot use L'Hopital's rule, in which case the conjugate method suggested by other answers would be more appropriate in this situation.

I attempted the problem, and I believe you can use L'Hopital's rule.

$$\lim_{x\to 1} \frac{(4x^2 +5)^{1/2} -3}{x^{2/3} -1}$$

This yields $\frac00$. So, using L'Hospitals:

$(4x^2 +5)^{1/2} -3$ becomes $(\frac12)(8x)(4x^2 +5)^{-1/2} = 4x(4x^2 +5)^{-1/2}$

$x^{2/3} -1$ becomes $(\frac23)(x^{-1/3})$

Now solve the limit:

$$\frac{4(1)(4(1)^2 +5)^{-1/2}}{(\frac23)(1^{-1/3})} = \frac{4(1)(1^{1/3})}{\frac23(4(1)^2 +5)^{1/2}} = \frac42 = 2$$

Hope this helped! (Apologies for lack of LaTex)

• I believe that L'Hopital rule can be really handfull in this problem, but it´s mandatory not to use it Commented Jan 19, 2018 at 17:58
• It can be solved with l'Hopital, yes. But I took the question to specify that he is not allowed to use l'Hopital. Commented Jan 19, 2018 at 17:58
• "and I believe you can use L'Hopital's rule" but OP clearly stated "I can´t use L'Hopital rule nor any derivatives." ...?! Commented Jan 19, 2018 at 17:58
• My apologies. I thought he tried but didn't succeed; I misread. Commented Jan 19, 2018 at 18:02
• I've added in MathJax($\LaTeX$) to the post but for future please look at this tutorial because it can be really hard to read answers without MathJax (also you can look at my edit to get a general idea). Commented Jan 19, 2018 at 18:16

$$\lim_{x\to 1} \frac{\sqrt{4x^2+5}-3}{\sqrt[3]{x^2}-1}=\lim_{x\rightarrow1}\frac{\sqrt{4x^6+5}-3}{x^2-1}=$$ $$=\lim_{x\rightarrow1}\frac{4x^6-4}{(x^2-1)(\sqrt{4x^6+5}+3)}=\lim_{x\rightarrow1}\frac{4(x^4+x^2+1)}{\sqrt{4x^6+5}+3}=2.$$

• Why can you do the first thing? Commented Jan 19, 2018 at 19:13
• It's the substitution $x\rightarrow x^3$. Also, it's obvious that $x\rightarrow1\Leftrightarrow x^3\rightarrow1.$ Commented Jan 19, 2018 at 19:14