Question

$$\lim _ { x \rightarrow \infty } \frac { \sqrt { x ^ { 4 } + 1 } - 2 x ^ { 2 } - 1 } { x ^ { 2 } }$$

Answer

$$l<-(c*(sqrt(x^4+1)-2*x^2-1))/(IM*x^3*m)$$

Solution


Use this rule: \(a \times \frac{b}{c}=\frac{ab}{c}\).
\[l\imath mx->\frac{c(\sqrt{{x}^{4}+1}-2{x}^{2}-1)}{{x}^{2}}\]
Regroup terms.
\[-+l\imath mx>\frac{c(\sqrt{{x}^{4}+1}-2{x}^{2}-1)}{{x}^{2}}\]
Divide both sides by \(\imath \).
\[-+lmx>\frac{\frac{c(\sqrt{{x}^{4}+1}-2{x}^{2}-1)}{{x}^{2}}}{\imath }\]
Simplify  \(\frac{\frac{c(\sqrt{{x}^{4}+1}-2{x}^{2}-1)}{{x}^{2}}}{\imath }\)  to  \(\frac{c(\sqrt{{x}^{4}+1}-2{x}^{2}-1)}{\imath {x}^{2}}\).
\[-+lmx>\frac{c(\sqrt{{x}^{4}+1}-2{x}^{2}-1)}{\imath {x}^{2}}\]
Divide both sides by \(m\).
\[-+lx>\frac{\frac{c(\sqrt{{x}^{4}+1}-2{x}^{2}-1)}{\imath {x}^{2}}}{m}\]
Simplify  \(\frac{\frac{c(\sqrt{{x}^{4}+1}-2{x}^{2}-1)}{\imath {x}^{2}}}{m}\)  to  \(\frac{c(\sqrt{{x}^{4}+1}-2{x}^{2}-1)}{\imath {x}^{2}m}\).
\[-+lx>\frac{c(\sqrt{{x}^{4}+1}-2{x}^{2}-1)}{\imath {x}^{2}m}\]
Divide both sides by \(x\).
\[-+l>\frac{\frac{c(\sqrt{{x}^{4}+1}-2{x}^{2}-1)}{\imath {x}^{2}m}}{x}\]
Simplify  \(\frac{\frac{c(\sqrt{{x}^{4}+1}-2{x}^{2}-1)}{\imath {x}^{2}m}}{x}\)  to  \(\frac{c(\sqrt{{x}^{4}+1}-2{x}^{2}-1)}{\imath {x}^{2}mx}\).
\[-+l>\frac{c(\sqrt{{x}^{4}+1}-2{x}^{2}-1)}{\imath {x}^{2}mx}\]
Regroup terms.
\[-+l>\frac{c(\sqrt{{x}^{4}+1}-2{x}^{2}-1)}{{x}^{2}xm\imath }\]
Use Product Rule: \({x}^{a}{x}^{b}={x}^{a+b}\).
\[-+l>\frac{c(\sqrt{{x}^{4}+1}-2{x}^{2}-1)}{{x}^{2+1}m\imath }\]
Simplify  \(2+1\)  to  \(3\).
\[-+l>\frac{c(\sqrt{{x}^{4}+1}-2{x}^{2}-1)}{{x}^{3}m\imath }\]
Regroup terms.
\[-+l>\frac{c(\sqrt{{x}^{4}+1}-2{x}^{2}-1)}{\imath {x}^{3}m}\]
Multiply both sides by \(-1\).
\[l<-\frac{c(\sqrt{{x}^{4}+1}-2{x}^{2}-1)}{\imath {x}^{3}m}\]