Rom

\(\displaystyle{\lim_{x \to \infty}}\dfrac{x^2+4x+3}{x+1}-(\alpha x + \beta)=3\)

\(\displaystyle{\lim_{x\to \infty}}\dfrac{(x+3)(x+1)}{x+1}-(\alpha x+\beta)\)

\(\displaystyle{\lim_{x\to \infty}}(x+3)-(\alpha x+\beta)=3\)

\(\displaystyle{\lim_{x\to \infty}}(1-\alpha)x+(3-\beta)=3\\ \\ \alpha = 1 \\ \beta = 0\)

\(\displaystyle{\lim_{x\to \infty}}\sqrt{4x^2-x}-\sqrt{9x^2+1}+x\)

\(\displaystyle{\lim_{x\to \infty}}x\left(\sqrt{4-\frac 1 x}-\sqrt{3+\frac 1 x}+1\right)\)

\(\mbox{In the limit we can ignore the }\dfrac 1 x \mbox{ terms}\)

\(\displaystyle{\lim_{x\to \infty}}x\left(\sqrt{4}-\sqrt{3}+1\right)=\displaystyle{\lim_{x\to \infty}}x\left(3-\sqrt{3}\right)\)

\(3 > \sqrt{3} \mbox{ so } \\ \displaystyle{\lim_{x\to \infty}}x\left(3-\sqrt{3}\right)=+\infty \\ \mbox{and so } \\ \displaystyle{\lim_{x\to \infty}}\sqrt{4x^2-x}-\sqrt{9x^2+1}+x=+\infty\)

\(x^2 + 2x\)

\(\mbox{note that }(x+a)^2 = x^2 + 2a x + a^2\)

\(\mbox{so we must have that }2a=2 \Rightarrow a=1\)

\(\mbox{and so we get } \\ x^2+2x = x^2+2x + 1 -1 = (x+1)^2 - 1\)

you could try 

96 and 1

or even

95 and 0

Assuming the total number of berries was some multiple of 5 you are correct.

\(410i = 16.40 \\ i = \dfrac {16.40}{410} \\ i = 0.04 \\ i = (0.04 \times 100)\% = 4\%\)

anything divided by 0, including 0, is undefined.

\(\left(5^8\right)^2 = 5^{8\cdot 2} = 5^{2\cdot 8} = \left(5^2\right)^8 = 25^8\)

You are correct

(4 + 2 + 5)k = 55

11k = 55

k =5

4x5 = 20 nickels = $1

2x5 = 10 dimes = $1

5x5 = 25 quarters = $6.25

total value = $8.25