Levi Rizki Saputra Notes

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lim⁡x→∞ax2+bx+c−px2+qx+r=lim⁡x→∞(ax2+bx+c−px2+qx+r)(ax2+bx+c+px2+qx+rax2+bx+c+px2+qx+r)=lim⁡x→∞(ax2+bx+c−px2+qx+r)(ax2+bx+c+px2+qx+r)ax2+bx+c+px2+qx+r=lim⁡x→∞(ax2+bx+c)2−(px2+qx+r)2ax2+bx+c+px2+qx+r=lim⁡x→∞ax2+bx+c−(px2+qx+r)ax2+bx+c+px2+qx+r=lim⁡x→∞ax2+bx+c−(px2+qx+r)ax2+bx+c+px2+qx+r((1x)(1x))=lim⁡x→∞(ax2+bx+c−(px2+qx+r)x)(ax2+bx+c+px2+qx+rx)=lim⁡x→∞ax + b +cx−(px+q+rx)ax2+bx+cx2+px2+qx+rx2=lim⁡x→∞ax + b +cx−(px+q+rx)a+bx+cx2+p+qx+rx2\begin{align*} &\lim _{x\rightarrow \infty }\sqrt{ax^{2} +bx+c} -\sqrt{px^{2} +qx+r}\\ &=\lim _{x\rightarrow \infty }\left(\sqrt{ax^{2} +bx+c} -\sqrt{px^{2} +qx+r}\right)\left(\dfrac{\sqrt{ax^{2} +bx+c} +\sqrt{px^{2} +qx+r}}{\sqrt{ax^{2} +bx+c} +\sqrt{px^{2} +qx+r}}\right)\\ &=\lim _{x\rightarrow \infty }\frac{\left(\sqrt{ax^{2} +bx+c} -\sqrt{px^{2} +qx+r}\right)\left(\sqrt{ax^{2} +bx+c} +\sqrt{px^{2} +qx+r}\right)}{\sqrt{ax^{2} +bx+c} +\sqrt{px^{2} +qx+r}}\\ &=\lim _{x\rightarrow \infty }\frac{\left(\sqrt{ax^{2} +bx+c}\right)^{2} -\left(\sqrt{px^{2} +qx+r}\right)^{2}}{\sqrt{ax^{2} +bx+c} +\sqrt{px^{2} +qx+r}}\\ &=\lim _{x\rightarrow \infty }\frac{ax^{2} +bx+c -\left( px^{2} +qx+r\right)}{\sqrt{ax^{2} +bx+c} +\sqrt{px^{2} +qx+r}}\\ &=\lim _{x\rightarrow \infty }\frac{ax^{2} +bx+c -\left( px^{2} +qx+r\right)}{\sqrt{ax^{2} +bx+c} +\sqrt{px^{2} +qx+r}}\left(\frac{\left(\frac{1}{x}\right)}{\left(\frac{1}{x}\right)}\right)\\ &=\lim _{x\rightarrow \infty }\frac{\left(\dfrac{ax^{2} +bx+c -\left( px^{2} +qx+r\right)}{x}\right)}{ \begin{array}{l} \left(\dfrac{\sqrt{ax^{2} +bx+c} +\sqrt{px^{2} +qx+r}}{x}\right)\\ \end{array}}\\ &=\lim _{x\rightarrow \infty }\dfrac{ax\ +\ b\ +\dfrac{c}{x} -\left( px+q+\dfrac{r}{x}\right)}{\sqrt{\dfrac{ax^{2} +bx+c}{x^{2}}} +\sqrt{\dfrac{px^{2} +qx+r}{x^{2}}}}\\ &=\lim _{x\rightarrow \infty }\frac{ax\ +\ b\ +\dfrac{c}{x} -\left( px+q+\dfrac{r}{x}\right)}{\sqrt{a+\dfrac{b}{x} +\dfrac{c}{x^{2}}} +\sqrt{p+\dfrac{q}{x} +\dfrac{r}{x^{2}}}} \end{align*}

Saat x→∞x\to\infty, 1x1 \over x dan 1x21 \over x^2 akan mendekati 0. Ini berarti konstanta yang dikalikan dengan 1x1 \over x dan 1x21 \over x^2 bisa kita abaikan.

lim⁡x→∞ax2+bx+c−px2+qx+r=lim⁡x→∞ax + b +cx−(px+q+rx)a+bx+cx2+p+qx+rx2=lim⁡x→∞ax + b −(px+q)a+p=lim⁡x→∞(a−p)x + b −qa+p \begin{align*} &\lim _{x\rightarrow \infty }\sqrt{ax^{2} +bx+c} -\sqrt{px^{2} +qx+r}\\ &=\lim _{x\rightarrow \infty }\frac{ax\ +\ b\ +\dfrac{c}{x} -\left( px+q+\dfrac{r}{x}\right)}{\sqrt{a+\dfrac{b}{x} +\dfrac{c}{x^{2}}} +\sqrt{p+\dfrac{q}{x} +\dfrac{r}{x^{2}}}}\\ &=\lim _{x\rightarrow \infty }\frac{ax\ +\ b\ -( px+q)}{\sqrt{a} +\sqrt{p}}\\ &=\lim _{x\rightarrow \infty }\frac{( a-p) x\ +\ b\ -q}{\sqrt{a} +\sqrt{p}} \end{align*}

Saat a=pa = p, maka a−p=0a-p=0 sehingga

lim⁡x→∞ax2+bx+c−px2+qx+r=lim⁡x→∞(a−p)x + b −qa+p=lim⁡x→∞ b −q2a= b −q2a \begin{align*} &\lim _{x\rightarrow \infty }\sqrt{ax^{2} +bx+c} -\sqrt{px^{2} +qx+r}\\ &=\lim _{x\rightarrow \infty }\frac{( a-p) x\ +\ b\ -q}{\sqrt{a} +\sqrt{p}}\\ \\ &=\lim _{x\rightarrow \infty }\frac{\ b\ -q}{2\sqrt{a}}\\ &=\frac{\ b\ -q}{2\sqrt{a}} \end{align*}

Saat a>ba \gt b, maka a−b>0a-b\gt 0 atau bilangan positif sehingga:

lim⁡x→∞ax2+bx+c−px2+qx+r=lim⁡x→∞(a−p)x + b −qa+p=(a−p)∞ + b −qa+p=∞ + b −qa+p=∞a+p=∞ \begin{align*} &\lim _{x\rightarrow \infty }\sqrt{ax^{2} +bx+c} -\sqrt{px^{2} +qx+r}\\ &=\lim _{x\rightarrow \infty }\frac{( a-p) x\ +\ b\ -q}{\sqrt{a} +\sqrt{p}}\\ \\ &=\frac{( a-p) \infty \ +\ b\ -q}{\sqrt{a} +\sqrt{p}}\\ &=\frac{\infty \ +\ b\ -q}{\sqrt{a} +\sqrt{p}}\\ &=\frac{\infty }{\sqrt{a} +\sqrt{p}}\\ &=\infty \\ \end{align*}

Saat a<ba \lt b, maka a−b<0a-b\lt 0 atau bilangan negatif sehingga:

lim⁡x→∞ax2+bx+c−px2+qx+r=lim⁡x→∞(a−p)x + b −qa+p=(a−p)∞ + b −qa+p=−∞ + b −qa+p=−∞a+p=−∞ \begin{align*} &\lim _{x\rightarrow \infty }\sqrt{ax^{2} +bx+c} -\sqrt{px^{2} +qx+r}\\ &=\lim _{x\rightarrow \infty }\frac{( a-p) x\ +\ b\ -q}{\sqrt{a} +\sqrt{p}}\\ \\ &=\frac{( a-p) \infty \ +\ b\ -q}{\sqrt{a} +\sqrt{p}}\\ &=\frac{-\infty \ +\ b\ -q}{\sqrt{a} +\sqrt{p}}\\ &=\frac{-\infty }{\sqrt{a} +\sqrt{p}}\\ &=-\infty \\ \end{align*}