find the limit of the following. hint: try multiplying and dividing by the conjugate. (lim_{x\rightarrowinfty…

Explanation:

Step1: Multiply and divide by conjugate

Multiply and divide $\lim_{x\rightarrow\infty}(\sqrt{4x^{2}+8x}-\sqrt{4x^{2}-7x})$ by $\sqrt{4x^{2}+8x}+\sqrt{4x^{2}-7x}$: [ \begin{align*} &\lim_{x\rightarrow\infty}\frac{(\sqrt{4x^{2}+8x}-\sqrt{4x^{2}-7x})(\sqrt{4x^{2}+8x}+\sqrt{4x^{2}-7x})}{\sqrt{4x^{2}+8x}+\sqrt{4x^{2}-7x}}\ =&\lim_{x\rightarrow\infty}\frac{(4x^{2}+8x)-(4x^{2}-7x)}{\sqrt{4x^{2}+8x}+\sqrt{4x^{2}-7x}} \end{align*} ]

Step2: Simplify the numerator

Simplify the numerator $(4x^{2}+8x)-(4x^{2}-7x)$: [ \begin{align*} (4x^{2}+8x)-(4x^{2}-7x)&=4x^{2}+8x - 4x^{2}+7x\ &=15x \end{align*} ] So we have $\lim_{x\rightarrow\infty}\frac{15x}{\sqrt{4x^{2}+8x}+\sqrt{4x^{2}-7x}}$.

Step3: Factor out $x^{2}$ from square - roots

Since $x\rightarrow\infty$, we can factor out $x^{2}$ from the square - roots in the denominator. $\sqrt{4x^{2}+8x}=x\sqrt{4 + \frac{8}{x}}$ and $\sqrt{4x^{2}-7x}=x\sqrt{4-\frac{7}{x}}$ (because $x>0$ as $x\rightarrow\infty$). The limit becomes $\lim_{x\rightarrow\infty}\frac{15x}{x\sqrt{4+\frac{8}{x}}+x\sqrt{4 - \frac{7}{x}}}$.

Step4: Cancel out $x$

Cancel out the common factor $x$ in the numerator and denominator: [ \begin{align*} \lim_{x\rightarrow\infty}\frac{15x}{x(\sqrt{4+\frac{8}{x}}+\sqrt{4 - \frac{7}{x}})}&=\lim_{x\rightarrow\infty}\frac{15}{\sqrt{4+\frac{8}{x}}+\sqrt{4 - \frac{7}{x}}} \end{align*} ]

Step5: Evaluate the limit

As $x\rightarrow\infty$, $\frac{8}{x}\rightarrow0$ and $\frac{7}{x}\rightarrow0$. [ \begin{align*} \lim_{x\rightarrow\infty}\frac{15}{\sqrt{4+\frac{8}{x}}+\sqrt{4 - \frac{7}{x}}}&=\frac{15}{\sqrt{4 + 0}+\sqrt{4-0}}\ &=\frac{15}{2 + 2}\ &=\frac{15}{4} \end{align*} ]

Answer:

$\frac{15}{4}$