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

Question

Can you please help me with this limit? I can´t use L'Hopital rule.

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

Answer 1

Make the ratio of the high degree terms which gives here $$ \frac{2x}{x^{4/3}} \underset{x \rightarrow +\infty}{\rightarrow}0 $$ EDIT :

I will propose my way of doing things. You are studying the limit around $1$ then, makes it move to $0$. Let $x=1+h$ $$ \frac{\sqrt{4\left(1+h\right)^2+5}-3}{\sqrt[3]{1+h}-1}=\frac{\sqrt{8h+4h^2+9}-3}{\sqrt[3]{1+h}-1} $$ First $$ \sqrt{8h+4h^2+9}=3\sqrt{1+\frac{8}{9}h+\frac{4}{9}h^2}=3+\frac{8}{18}h+o\left(h\right) $$ and $$ \sqrt[3]{1+h}-1=1+\frac{h}{3}-1+o\left(h\right)=\frac{h}{3}+o\left(h\right) $$ So

$$ \frac{\sqrt{4\left(1+h\right)^2+5}-3}{\sqrt[3]{1+h}-1}\underset{(0)}{=}\frac{3+\frac{8}{18}h+o\left(h\right)-3}{\frac{h}{3}+o\left(h\right)}\underset{h \rightarrow 0}{\rightarrow}\frac{24}{18}=\frac{4}{3} $$

Then we conclude that

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

Answer 2

Since the OP seems to be doubtfully on the whether $x\to \infty$ or $x\to1$ I have included both answers:

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If $x\to \infty$ then we have $$\lim_{x\to \infty} \frac{\sqrt{4x^2+5}-3}{\sqrt[3]{x^4}-1} =\lim_{x\to \infty}{\sqrt{4+{5\over x^2}}-{3\over x}\over \sqrt[3]{x}-{1\over x}} =\lim_{x\to \infty}{\sqrt{4}\over \sqrt[3]{x}} = 0$$

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And, if $x\to 1,$ then we have:

By definition of derivative at $x=1$ we have $$\lim_{x\to 1} \frac{\sqrt{4x^2+5}-3}{\sqrt[3]{x^4}-1} = \lim_{x\to 1} \frac{\sqrt{4x^2+5}-3}{x-1}\lim_{x\to 1} \frac{x-1}{\sqrt[3]{x^4}-1} = \left(\sqrt{4x^2+5}\right)'\frac{1}{\left(\sqrt[3]{x^4}\right)'}\Bigg|_{x=1} = 1$$

Since $$\left(\sqrt{4x^2+5}\right)'\Bigg|_{x=1} = \frac{4}{\sqrt{4x^2+5}}\Bigg|_{x=1} = \frac{4}{3}$$

and $$\left(\sqrt[3]{x^4}\right)'\Bigg|_{x=1} = \frac{4}{3}x^{4/3-1}\Bigg|_{x=1} = \frac{4}{3}$$

Answer 3

Rewrite the fraction as follows

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

And this has $0$ as limit at $+\infty$

At $1$ we have a $0/0$ indetermination. Many ways to solve, the most elementary being the conjugate radicals.

Multiply both numerator and denominator by $\sqrt{4x^2+5}+3$. This yields

$${4(x^2-1)\over \sqrt[3]{x^4}-1}\cdot{1\over \sqrt{4x^2+5}+3}$$

We still have a $0/0$ indetermination in the first term of the product while the second has $1\over 6$ as limit at $1$.

To solve the indetermination we’re left with we need to use the identity $a^3-b^3=(a-b)(a^2+ab+b^2)$. So we multiply both numerator and denominator by $(\sqrt[3]{x^4})^2+\sqrt[3]{x^4}+1$ to get:

$${4(x^2-1)\over \sqrt[3]{x^4}-1}={4(x^2-1)\cdot((\sqrt[3]{x^4})^2+\sqrt[3]{x^4}+1)\over x^4-1}=4{(\sqrt[3]{x^4})^2+\sqrt[3]{x^4}+1\over x^2+1}$$

And for this term the limit is $6$ when $x\to 1$ and so the limit we’re looking for is $6/6=1$

Answer 4

Hint: use that $$a^2 - b^2 = (a - b)(a + b)$$ and $$a^3 - b^3 = (a - b)(a^2 + ab + b^2).$$