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