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

Question

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.

Answer 1

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,

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

Answer 2

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

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

we obtain

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

Answer 3

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

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 $\sqrt[3]{x}-1$.

Answer 4

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)

Answer 5

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