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

Question

find limits :

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

my try :

$\lim_{x\rightarrow\infty} (\sqrt[p]{x^{n}+a_{n}x^{n-1}+\cdots+a_{0}}- \sqrt[q]{x^{m}+a_{m}x^{m-1}+\cdots+a_{0}})=?$ $$\lim_{ x \to \infty }\sqrt[2]{x^4+3x^2+5}-\sqrt[2]{x^4-5x^2+1}=L$$

$$L=x^{\frac{n}{p}}-x^{\frac{m}{q}}+\frac{a_n}{p}x^{\frac{n}{p}-1}-\frac{a_m}{q}x^{\frac{m}{q}-1}+o\left(x^{\frac{n}{p}-1}\right)+o\left(x^{\frac{m}{q}-1}\right)$$

$$L=x^{\frac{4}{2}}-x^{\frac{4}{2}}+\frac{3}{2}x^{\frac{4}{2}-1}-\frac{(-5)}{2}x^{\frac{4}{2}-1}+o\left(x^{\frac{n}{p}-1}\right)+o\left(x^{\frac{m}{q}-1}\right) \\=4x+o\left(x^{\frac{n}{p}-1}\right)+o\left(x^{\frac{m}{q}-1}\right) \\=\infty?$$

Answer 1

Standard technique: $$ \sqrt{x^4+3x^2+5}-\sqrt{x^4-5x^2+1}= \frac{x^4+3x^2+5-x^4+5x^2-1} {\sqrt{x^4+3x^2+5}+\sqrt{x^4-5x^2+1}} $$ Simplify and divide numerator and denominator by $x^2$.

Answer 2

Hint: we have $$\sqrt{x^4+3x^2+5}-\sqrt{x^4-5x^2+1}=\frac{x^4+3x^2+5-x^4+5x^2-1}{\sqrt{x^4+3x^2+5}+\sqrt{x^4-5x^2+1}}$$

Answer 3

The usual trick to handling limits like $\lim_{x\to\infty} \sqrt{f(x)}-\sqrt{g(x)}$ is to rewrite it as:

$$\lim_{x\to\infty}\frac{f(x)-g(x)}{\sqrt{f(x)}+\sqrt{g(x)}}$$

In this case, $f(x)-g(x)=8x^2+4$ and $\sqrt{f(x)}$. Dividing numerator and denominator by $x^2$, then you get:

$$\lim_{x\to\infty}\frac{8+\frac{4}{x^2}}{\sqrt{1+\frac{3}{x^2}+\frac{5}{x^4}}+\sqrt{1-\frac{5}{x^2}+\frac{1}{x^4}}}=4$$

Answer 4

Since $\infty-\infty$ is an indefinite form, we need a large-$x$ approximation for the function whose limit is sought. Since $\sqrt{x^4+ax^2+b}=x^2\sqrt{1+ax^{-2}+bx^{-4}}\approx x^2+\frac{a}{2}$, the limit is $\frac{3}{2}-\frac{-5}{2}=4$.

Answer 5

Generalisation

For $t>0$, $m,n,k\in\mathbb{Z}_{>0}$, and $a_0,a_1,\ldots,a_{k(m-1)},b_0,b_1,\ldots,b_{k(n-1)}\in\mathbb{R}$, we set $$L:=\lim_{x\to\infty}\,\left(\sqrt[m]{t^mx^{km}+\sum_{i=0}^{k(m-1)}\,a_ix^i}-\sqrt[n]{t^nx^{kn}+\sum_{j=0}^{k(m-1)}\,b_jx^j}\right)\,.$$ Then, $$L=\lim_{x\to\infty}\,\left(\Bigg(tx^k+\frac{a_{k(m-1)}}{mt^{m-1}}+\mathcal{O}\left(\frac{1}{x}\right)\Bigg)^{\vphantom{a^a}}-\Bigg(tx^k+\frac{b_{k(n-1)}}{nt^{n-1}}+\mathcal{O}\left(\frac{1}{x}\right)\Bigg)\right)\,.$$ Thus, $$L=\frac{a_{k(m-1)}}{mt^{m-1}}-\frac{b_{k(n-1)}}{nt^{n-1}}\,.$$ In this problem, $m=n=k=2$, $t=1$, $a_2=3$, and $b_2=-5$, so $$L=\frac{3-(-5)}{2}=4\,.$$

Answer 6

Your broad approach should be all right, but you have to be more careful in the specific details. Note that we have $x^4+3x^2+5=x^4(1+\dfrac3{x^2}+\dfrac5{x^4})$, so $\sqrt{x^4+3x^2+5}=\sqrt{x^4}\sqrt{1+\dfrac3{x^2}+\dfrac5{x^4}}$ $=x^2(1+\dfrac3{2x^2}+O(x^{-4}))$ (where the last equation follows from the binomial expansion $(1+t)^a = 1+at+O(t^2)$). Can you figure out where to go from there?

Answer 7

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

put $x^2 = y$. As $x\to \infty ,$ so does $x^2 = y \to \infty$.

This gives us $$\lim_{y \to \infty }\sqrt[2]{y^2+3y+5}-\sqrt[2]{y^2-5y+1}$$

Now, multiply top and bottom by the conjugate. (E.g., the conjugate of $\sqrt{a}-\sqrt b$ is $\sqrt a +\sqrt b)$.

$$\lim_{y\to \infty} \frac{\left(\sqrt{y^2+3y+5}-\sqrt{y^2-5y+1}\right)+\left(\sqrt{y^2+3y+5}+\sqrt{y^2 -5y+1}\right)}{\sqrt{y^2+3y+5}+\sqrt{y^2 -5y+1}}$$

$$= \lim_{y\to \infty} \frac{\left({y^2+3y+5}-{(y^2-5y+1)}\right)}{\sqrt{y^2+3y+5}+\sqrt{y^2 -5y+1}}= \lim_{y\to \infty} \frac{8y+4}{\sqrt{y^2+3y+5}+\sqrt{y^2 -5y+1}}$$

$$=\lim_{y\to \infty} \frac{{y}\left( 8 +\frac 4{y}\right)}{y\Big(\sqrt{1+\frac 3y+\frac 5{y^2}} +\sqrt{1 -\frac 5y+\frac 1{y^2}}\Big)} = \frac 8{1+1} = 4$$