+198 I was just reviewing my calc book when I stumbled across this problem.
Any thoughts?
\(\lim_{x\longrightarrow\infty}\frac{\sqrt{x^4-10x}-\sqrt{x^4-5x^2+7}}{5}\)
+118609 \lim_{x\longrightarrow\infty}\frac{\sqrt{x^4-10x}-\sqrt{x^4-5x^2+7}}{5}
\(\displaystyle\lim_{x\rightarrow\infty}\;\frac{\sqrt{x^4-10x}-\sqrt{x^4-5x^2+7}}{5}\\ =\displaystyle\lim_{x\rightarrow\infty}\;\frac{ \left(\sqrt{x^4-10x}-\sqrt{x^4-5x^2+7}\right) \left(\sqrt{x^4-10x}+\sqrt{x^4-5x^2+7}\right)}{5 \left(\sqrt{x^4-10x}+\sqrt{x^4-5x^2+7}\right)}\\ =\displaystyle\lim_{x\rightarrow\infty}\;\frac{ x^4-10x-(x^4-5x^2+7) }{5 \left(\sqrt{x^4-10x}+\sqrt{x^4-5x^2+7}\right)}\\ =\displaystyle\lim_{x\rightarrow\infty}\;\frac{ 5x^2 -10x-7 }{5 \left(\sqrt{x^4-10x}+\sqrt{x^4-5x^2+7}\right)}\\ =\displaystyle\lim_{x\rightarrow\infty}\;\frac{ 5x^2 -10x-7 }{5 \left(\sqrt{x^4-10x}\right)\left(1+\frac{\sqrt{x^4-5x^2+7}}{\sqrt{x^4-10x}}\right)}\\ \)
\( =\displaystyle\frac{1}{5}\times\displaystyle\lim_{x\rightarrow\infty}\;\frac{ 5x^2 -10x-7 }{ \left(\sqrt{x^4-10x}\right)} \times\;\;\displaystyle\lim_{x\rightarrow\infty}\;\frac{ 1}{\left(1+\frac{\sqrt{x^4-5x^2+7}}{\sqrt{x^4-10x}}\right)}\\ \)
NOW I will look at each of these limits seperately.
\(\displaystyle\lim_{x\rightarrow\infty}\;\frac{ 5x^2 -10x-7 }{ \left(\sqrt{x^4-10x}\right)}\\ =\displaystyle\lim_{x\rightarrow\infty}\;\sqrt{\frac{ (5x^2 -10x-7 )^2 }{ x^4-10x}}\\ \text{expanding gives}\\ =\displaystyle\lim_{x\rightarrow\infty}\;\sqrt{\frac{ 25x^4-100x^3+30x^2+140x+49 }{ x^4-10x}}\\ \text{Dividing top and bottom by x^4 we get}\\ =\displaystyle\lim_{x\rightarrow\infty}\;\sqrt{\frac{ 25-\frac{100}{x}+\frac{30}{x^2}+\frac{140}{x^3}+\frac{49}{x^4} }{ 1-\frac{10}{x^3}}}\\ =\sqrt{25}\\ =5\\ \text{-------------------------------------------------------}\)
\(\displaystyle\lim_{x\rightarrow\infty}\;\frac{ 1}{\left(1+\frac{\sqrt{x^4-5x^2+7}}{\sqrt{x^4-10x}}\right)}\\~\\ =\frac{ 1}{\displaystyle\lim_{x\rightarrow\infty}\;\left(1+\frac{\sqrt{x^4-5x^2+7}}{\sqrt{x^4-10x}}\right)}\\ =\frac{ 1}{\displaystyle\lim_{x\rightarrow\infty}\;\left(\frac{\sqrt{x^4-5x^2+7}}{\sqrt{x^4-10x}}\right)+1}\\ =\frac{ 1}{\displaystyle\lim_{x\rightarrow\infty}\;\sqrt{\left(\frac{x^4-5x^2+7}{x^4-10x}\right)}+1}\\ =\frac{ 1}{\displaystyle\lim_{x\rightarrow\infty}\;\sqrt{\left(\frac{1-\frac{5}{x^2}+\frac{7}{x^4}}{1-\frac{10}{x^3}}\right)}+1}\\ =\frac{ 1}{\displaystyle\lim_{x\rightarrow\infty}\;\sqrt{\left(\frac{1-\frac{5}{x^2}+\frac{7}{x^4}}{1-\frac{10}{x^3}}\right)}+1}\\ =\displaystyle \frac{1}{2}\)
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SO what do we have now.
\(\displaystyle\lim_{x\rightarrow\infty}\;\frac{\sqrt{x^4-10x}-\sqrt{x^4-5x^2+7}}{5}= \frac{1}{5} \times5\times \frac{1}{2}=\boxed{\frac{1}{2}}\)
+26367 Limits
I was just reviewing my calc book when I stumbled across this problem.
\(\large { \lim \limits_{x \to \infty} \dfrac{\sqrt{x^4-10x}-\sqrt{x^4-5x^2+7}}{5} }\)
\(\begin{array}{|rcll|} \hline &&\mathbf{ \lim \limits_{x \to \infty} \dfrac{\sqrt{x^4-10x}-\sqrt{x^4-5x^2+7}}{5} } \\\\ &=&\dfrac15 \cdot \lim \limits_{x \to \infty} \left( \sqrt{x^4-10x}-\sqrt{x^4-5x^2+7} \right) \\\\ &=&\dfrac15 \cdot \lim \limits_{x \to \infty} \left( \sqrt{x^4-10x}-\sqrt{x^4-5x^2+7} \right) \dfrac{\left( \sqrt{x^4-10x}+\sqrt{x^4-5x^2+7} \right)}{\left( \sqrt{x^4-10x}+\sqrt{x^4-5x^2+7} \right)} \\\\ &=&\dfrac15 \cdot \lim \limits_{x \to \infty} \left( \dfrac{ x^4-10x-(x^4-5x^2+7)} { \sqrt{x^4-10x}+\sqrt{x^4-5x^2+7} } \right) \\\\ &=&\dfrac15 \cdot \lim \limits_{x \to \infty} \left( \dfrac{ 5x^2-10x-7} { \sqrt{x^4-10x}+\sqrt{x^4-5x^2+7} } \right) \cdot \dfrac{x^2}{x^2} \\\\ &=&\dfrac15 \cdot \lim \limits_{x \to \infty} \left( \dfrac{ \frac{5x^2-10x-7} {x^2} } { \frac{\sqrt{x^4-10x}+\sqrt{x^4-5x^2+7}} {x^2} } \right) \\\\ &=&\dfrac15 \cdot \lim \limits_{x \to \infty} \left( \dfrac{ \frac{5x^2-10x-7} {x^2} } { \sqrt{\frac{x^4-10x} {x^4} }+\sqrt{ \frac{x^4-5x^2+7} {x^4} } } \right) \\\\ &=&\dfrac15 \cdot \lim \limits_{x \to \infty} \left( \dfrac{ 5-\frac{10}{x}-\frac{7}{x^2} } { \sqrt{1-\frac{10}{x^3} } + \sqrt{1-\frac{5}{x^2}+\frac{7}{x^4}} } \right) \\\\ &=&\dfrac15 \cdot \left( \dfrac{ 5-0-0 } { \sqrt{1-0 } + \sqrt{1-0+0} } \right) \\\\ &=&\dfrac15 \cdot \left( \dfrac{ 5 } { 1 + 1 } \right) \\\\ &=&\dfrac15 \cdot \dfrac{ 5 } {2 } \\\\ &\mathbf{=}&\mathbf{\dfrac12} \\ \hline \end{array}\)
