Question

\[\lim_{x \to - \infty} \left( \sqrt{x^2 - 8x} + x \right)\] 

Answer 1

\[\lim_{x \to - \infty} \left( \sqrt{x^2 - 8x} + x \right)\] Let x = –m When x → –∞, then m → ∞.  

\[\lim_{m \to \infty} \left( \sqrt{m^2 + 8m} - m \right)\]
\[ = \lim_{m \to \infty} \left( \frac{\left( \sqrt{m^2 + 8m} - m \right) \left( \sqrt{m^2 + 8m} + m \right)}{\left( \sqrt{m^2 + 8m} + m \right)} \right)\]
\[ = \lim_{m \to \infty} \left[ \frac{m^2 + 8m - m^2}{\sqrt{m^2 + 8m} + m} \right]\] 

Dividing the numerator and the denominator by m: 

\[\lim_{m \to \infty} \left[ \frac{8}{\frac{\sqrt{m^2 + 8m} + 1}{m}} \right]\]
\[ = \lim_{m \to \infty} \left[ \frac{8}{\sqrt{\frac{m^2}{m^2} + \frac{8m}{m^2}} + 1} \right]\]
\[ = \lim_{m \to \infty} \left[ \frac{8}{\sqrt{1 + \frac{8}{m}} + 1} \right]\]
\[\text{ As } m \to \infty , \frac{1}{m} \to 0\]
\[ = \frac{8}{\sqrt{1 + 0} + 1}\]
\[ = 4\]