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I am having trouble solving this. Need to find the value of a so that the equation will equal $7$ when $x$ goes to infinity.

$$\lim_{x \rightarrow\infty} \left (\sqrt{x^2 + x} - \sqrt{x^2+ax} \right ) = 7$$

What I've tried is to multiply and divide the left side with the conjugate. I've also tried to multiply both sides with the conjugate.

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Yes your way is right and we obtain

$$\sqrt{x^2+x}-\sqrt{x^2+ax}=(\sqrt{x^2+x}-\sqrt{x^2+ax})\cdot\frac{\sqrt{x^2+x}+\sqrt{x^2+ax}}{\sqrt{x^2+x}+\sqrt{x^2+ax}}=\\=\frac{x-ax}{\sqrt{x^2+x}+\sqrt{x^2+ax}}$$

then factor out an $x$ term from numerator and denominator and take the limit.

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$ \sqrt {{x^{2}+x}} -\sqrt{{x^{2}+ax}}$$=\frac {(1-a)x} {\sqrt {{x^{2}+x}} +\sqrt{{x^{2}+ax}}}$. Divide throughout by $x$ to see th at the limit is $\frac {1-a} 2$. Hence you want $\frac {1-a} 2=7$ or $a=-13$.

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Set $1/x=h^2$

$$\lim_{x\to\infty}\sqrt{x^2+x}+\sqrt{x^2+ax}=\lim_{h\to0}\dfrac{\sqrt{1+h^2}-\sqrt{1+ah^2}}{h^2}$$

$$=\lim_{h\to0}\dfrac{{1+h^2}-(1+ah^2)}{h^2}\cdot\lim_{h\to0}\dfrac1{\sqrt{1+h^2}+\sqrt{1+ah^2}}=\dfrac{1-a}{1+1}$$

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