6.10. $$ \lim _{v \rightarrow -\infty}\left(\frac{2-v}{\sqrt{7+6 v^{2}}}\right) $$.

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Calculate the limit $$ \lim_{vightarrow -\infty}\frac{2-v}{\sqrt{7+6v^{2}}} $$.
Evaluate the limit by following steps:
- step0: Evaluate using transformations:
  $$\lim _{vightarrow -\infty}\left(\frac{2-v}{\sqrt{7+6v^{2}}}ight)$$
- step1: Rewrite the expression:
  $$\lim _{vightarrow -\infty}\left(\frac{\left(\frac{2}{v}-1ight)v}{-\sqrt{\frac{7}{v^{2}}+6}	imes v}ight)$$
- step2: Reduce the fraction:
  $$\lim _{vightarrow -\infty}\left(\frac{\frac{2}{v}-1}{-\sqrt{\frac{7}{v^{2}}+6}}ight)$$
- step3: Rewrite the expression:
  $$\frac{\lim _{vightarrow -\infty}\left(\frac{2}{v}-1ight)}{\lim _{vightarrow -\infty}\left(-\sqrt{\frac{7}{v^{2}}+6}ight)}$$
- step4: Calculate:
  $$\frac{-1}{\lim _{vightarrow -\infty}\left(-\sqrt{\frac{7}{v^{2}}+6}ight)}$$
- step5: Calculate:
  $$\frac{-1}{-\sqrt{6}}$$
- step6: Reduce the fraction:
  $$\frac{1}{\sqrt{6}}$$
- step7: Multiply by the Conjugate:
  $$\frac{\sqrt{6}}{\sqrt{6}	imes \sqrt{6}}$$
- step8: Multiply the numbers:
  $$\frac{\sqrt{6}}{6}$$
The limit of $$ \lim _{v ightarrow -\infty}\left(\frac{2-v}{\sqrt{7+6 v^{2}}}ight) $$ as $$ v $$ approaches negative infinity is $$ \frac{\sqrt{6}}{6} $$.
The limit is $$ \frac{\sqrt{6}}{6} $$.

6.10. .

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