Question

find the limit. (if the limit is infinite, enter ∞ or -∞, as appropriate. if the limit does not
\\( \lim _ { x \
ightarrow - \infty } \left( \sqrt { 49 x ^ { 2 } + 8 x } + 7 x \
ight) \\)

Explanation:

Step1: Rationalize the expression

Multiply and divide by the conjugate:

$$\lim_{x ightarrow-\infty}\frac{(\sqrt{49x^{2}+8x}+7x)(\sqrt{49x^{2}+8x}-7x)}{\sqrt{49x^{2}+8x}-7x}$$

Using the difference of squares \((a + b)(a - b)=a^{2}-b^{2}\), we get:

$$\lim_{x ightarrow-\infty}\frac{49x^{2}+8x-(49x^{2})}{\sqrt{49x^{2}+8x}-7x}=\lim_{x ightarrow-\infty}\frac{8x}{\sqrt{49x^{2}+8x}-7x}$$

Step2: Simplify the denominator

Since \(x
ightarrow-\infty\), \(\sqrt{49x^{2}}=-7x\) (because \(x<0\)). Factor out \(x^{2}\) from the square - root:

$$\lim_{x ightarrow-\infty}\frac{8x}{|x|\sqrt{49+\frac{8}{x}}-7x}=\lim_{x ightarrow-\infty}\frac{8x}{-x\sqrt{49+\frac{8}{x}}-7x}$$

Factor out \(x\) from the denominator:

$$\lim_{x ightarrow-\infty}\frac{8x}{x(-\sqrt{49+\frac{8}{x}}-7)}$$

Cancel out \(x\) (since \(x
eq0\) as \(x
ightarrow-\infty\)):

$$\lim_{x ightarrow-\infty}\frac{8}{-\sqrt{49+\frac{8}{x}}-7}$$

Step3: Evaluate the limit

As \(x
ightarrow-\infty\), \(\frac{8}{x}
ightarrow0\).
$$\frac{8}{-\sqrt{49 + 0}-7}=\frac{8}{-7 - 7}=-\frac{8}{14}$$

Answer:

\(-\frac{8}{14}\)