Caculus 1 limit at infinity
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Try the substitution \(\displaystyle y = \sqrt{x}.\)limit (3x^3+sqrt(x)-1)/(4x^5+3x+1) as x->infinity
I'm not sure how to do this? theis throwing me off
D
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limit (3x^3+sqrt(x)-1)/(4x^5+3x+1) as x->infinity
I'm not sure how to do this? theis throwing me off
Divide numerator and the denominators by x^5 - and take the limit.
Hello, andy849!
What would you do if the \(\displaystyle \sqrt{x}\) wasn't there?
Divide numerator and denominator by \(\displaystyle x^5\)
. . \(\displaystyle \displaystyle \lim_{x\to\infty}\frac{\frac{3x^3}{x^5} + \frac{x^{\frac{1}{2}}}{x^5} - \frac{1}{x^5}} {\frac{4x^5}{x^5} + \frac{3x}{x^5} + \frac{1}{x^5}} \;=\;\lim_{x\to\infty}\frac{\frac{3}{x^2} + \frac{1}{x^{9/2}} - \frac{1}{x^5}}{4 + \frac{3}{x^4} + \frac{1}{x^5}} \;=\;\frac{0+0-0}{4+0+0} \;=\;0\)
What would you do if the \(\displaystyle \sqrt{x}\) wasn't there?
Divide numerator and denominator by \(\displaystyle x^5\)
. . \(\displaystyle \displaystyle \lim_{x\to\infty}\frac{\frac{3x^3}{x^5} + \frac{x^{\frac{1}{2}}}{x^5} - \frac{1}{x^5}} {\frac{4x^5}{x^5} + \frac{3x}{x^5} + \frac{1}{x^5}} \;=\;\lim_{x\to\infty}\frac{\frac{3}{x^2} + \frac{1}{x^{9/2}} - \frac{1}{x^5}}{4 + \frac{3}{x^4} + \frac{1}{x^5}} \;=\;\frac{0+0-0}{4+0+0} \;=\;0\)
Given that soroban has shown you how to do the limit process, I’ll show you how the substitution process leads to the exact same result.
\(\displaystyle y = \sqrt{x}\implies \dfrac{3x^3+\sqrt{x}-1}{4x^5+3x+1} = \dfrac{3y^6+y-1}{4y^{10} + 3y^2 +1}.\) Just an ordinary rational function
\(\displaystyle \displaystyle \lim_{y \rightarrow \infty} \dfrac{3y^6+y-1}{4y^{10}+3y^2+1} = \lim_{y \rightarrow \infty}\dfrac{\frac{3}{y^4}+\frac{1}{y^9}-\frac{1}{y^{10}}}{4+\frac{3}{y^8}+\frac{1}{y^{10}}}.\) Divided both numerator and denominator by y^10..
\(\displaystyle So\ \displaystyle \lim_{y \rightarrow \infty} \dfrac{3y^6+y-1}{4y^{10}+3y^2+1} = \dfrac{0 + 0 + 0}{4 + 0 + 0} = 0.\) Same as soroban.
When something baffles you, try to get rid of it with a substitution.
\(\displaystyle y = \sqrt{x}\implies \dfrac{3x^3+\sqrt{x}-1}{4x^5+3x+1} = \dfrac{3y^6+y-1}{4y^{10} + 3y^2 +1}.\) Just an ordinary rational function
\(\displaystyle \displaystyle \lim_{y \rightarrow \infty} \dfrac{3y^6+y-1}{4y^{10}+3y^2+1} = \lim_{y \rightarrow \infty}\dfrac{\frac{3}{y^4}+\frac{1}{y^9}-\frac{1}{y^{10}}}{4+\frac{3}{y^8}+\frac{1}{y^{10}}}.\) Divided both numerator and denominator by y^10..
\(\displaystyle So\ \displaystyle \lim_{y \rightarrow \infty} \dfrac{3y^6+y-1}{4y^{10}+3y^2+1} = \dfrac{0 + 0 + 0}{4 + 0 + 0} = 0.\) Same as soroban.
When something baffles you, try to get rid of it with a substitution.
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