Limits Asymptotes 1C8F66

1. **Problem statement:** Calculate the following limits and find the asymptotes of given curves. --- **(i) Calculate** $$\lim_{x \to \pm \infty} \frac{2x + \sqrt{x^2 + 1}}{x - 2}$$ 2. For large $|x|$, factor $x$ from the square root: $$\sqrt{x^2 + 1} = |x|\sqrt{1 + \frac{1}{x^2}}$$ 3. As $x \to +\infty$, $|x| = x$, so: $$\frac{2x + x\sqrt{1 + \frac{1}{x^2}}}{x - 2} = \frac{x(2 + \sqrt{1 + \frac{1}{x^2}})}{x(1 - \frac{2}{x})} = \frac{2 + \sqrt{1 + \frac{1}{x^2}}}{1 - \frac{2}{x}}$$ 4. Taking limit $x \to +\infty$, terms with $\frac{1}{x}$ vanish: $$= \frac{2 + 1}{1 - 0} = 3$$ 5. As $x \to -\infty$, $|x| = -x$, so: $$\frac{2x - x\sqrt{1 + \frac{1}{x^2}}}{x - 2} = \frac{x(2 - \sqrt{1 + \frac{1}{x^2}})}{x(1 - \frac{2}{x})} = \frac{2 - \sqrt{1 + \frac{1}{x^2}}}{1 - \frac{2}{x}}$$ 6. Taking limit $x \to -\infty$: $$= \frac{2 - 1}{1 - 0} = 1$$ **Answer:** $$\lim_{x \to +\infty} = 3, \quad \lim_{x \to -\infty} = 1$$ --- **(j) Calculate** $$\lim_{x \to \pm \infty} \sqrt{x^2 + 4} - x$$ 2. Factor $x^2$ inside the root: $$\sqrt{x^2 + 4} = |x|\sqrt{1 + \frac{4}{x^2}}$$ 3. For $x \to +\infty$, $|x| = x$, so: $$\sqrt{x^2 + 4} - x = x\sqrt{1 + \frac{4}{x^2}} - x = x\left(\sqrt{1 + \frac{4}{x^2}} - 1\right)$$ 4. Use binomial expansion for large $x$: $$\sqrt{1 + \frac{4}{x^2}} \approx 1 + \frac{2}{x^2}$$ 5. So: $$x\left(1 + \frac{2}{x^2} - 1\right) = x \cdot \frac{2}{x^2} = \frac{2}{x} \to 0$$ 6. For $x \to -\infty$, $|x| = -x$, so: $$\sqrt{x^2 + 4} - x = -x\sqrt{1 + \frac{4}{x^2}} - x = -x\left(\sqrt{1 + \frac{4}{x^2}} + 1\right)$$ 7. As $x \to -\infty$, $-x \to +\infty$, and $\sqrt{1 + \frac{4}{x^2}} + 1 \to 2$, so limit diverges to $+\infty$. **Answer:** $$\lim_{x \to +\infty} = 0, \quad \lim_{x \to -\infty} = +\infty$$ --- **(k) Calculate** $$\lim_{x \to \pm \infty} \sqrt{3x^2 + 1} - \sqrt{2x^2 + x}$$ 2. Rationalize by multiplying numerator and denominator by the conjugate: $$\frac{(\sqrt{3x^2 + 1} - \sqrt{2x^2 + x})(\sqrt{3x^2 + 1} + \sqrt{2x^2 + x})}{\sqrt{3x^2 + 1} + \sqrt{2x^2 + x}} = \frac{3x^2 + 1 - (2x^2 + x)}{\sqrt{3x^2 + 1} + \sqrt{2x^2 + x}}$$ 3. Simplify numerator: $$3x^2 + 1 - 2x^2 - x = x^2 + 1 - x$$ 4. Divide numerator and denominator by $|x|$: $$\frac{x^2 + 1 - x}{\sqrt{3x^2 + 1} + \sqrt{2x^2 + x}} = \frac{x^2(1 + \frac{1}{x^2} - \frac{1}{x})}{|x|(\sqrt{3 + \frac{1}{x^2}} + \sqrt{2 + \frac{1}{x}})} = \frac{|x| (1 + \frac{1}{x^2} - \frac{1}{x})}{\sqrt{3 + \frac{1}{x^2}} + \sqrt{2 + \frac{1}{x}}}$$ 5. For $x \to +\infty$, $|x| = x$, so: $$\lim_{x \to +\infty} = \lim_{x \to +\infty} \frac{x(1 + 0 - 0)}{\sqrt{3 + 0} + \sqrt{2 + 0}} = \frac{x}{\sqrt{3} + \sqrt{2}} \to +\infty$$ 6. For $x \to -\infty$, $|x| = -x$, so numerator becomes: $$-x(1 + 0 - 0) = -x$$ 7. Denominator: $$\sqrt{3 + 0} + \sqrt{2 + 0} = \sqrt{3} + \sqrt{2}$$ 8. So limit: $$\lim_{x \to -\infty} = \frac{-x}{\sqrt{3} + \sqrt{2}} \to -\infty$$ **Answer:** $$\lim_{x \to +\infty} = +\infty, \quad \lim_{x \to -\infty} = -\infty$$ --- **(l) Calculate** $$\lim_{x \to -5} \frac{x^3 + 5x^2 + x + 5}{x^2 + 12x + 35}$$ 2. Factor denominator: $$x^2 + 12x + 35 = (x + 5)(x + 7)$$ 3. Factor numerator by grouping: $$x^3 + 5x^2 + x + 5 = x^2(x + 5) + 1(x + 5) = (x + 5)(x^2 + 1)$$ 4. Cancel common factor $(x + 5)$: $$\frac{\cancel{(x + 5)}(x^2 + 1)}{\cancel{(x + 5)}(x + 7)} = \frac{x^2 + 1}{x + 7}$$ 5. Substitute $x = -5$: $$\frac{(-5)^2 + 1}{-5 + 7} = \frac{25 + 1}{2} = \frac{26}{2} = 13$$ **Answer:** $$13$$ --- **(m) Calculate** $$\lim_{x \to -1} \frac{x^2 + 3x + 2}{2x + 1 + \sqrt{3x + 4}}$$ 2. Factor numerator: $$x^2 + 3x + 2 = (x + 1)(x + 2)$$ 3. Substitute $x = -1$ in denominator: $$2(-1) + 1 + \sqrt{3(-1) + 4} = -2 + 1 + \sqrt{-3 + 4} = -1 + 1 = 0$$ 4. Substitute $x = -1$ in numerator: $$( -1 + 1)(-1 + 2) = 0 \cdot 1 = 0$$ 5. Indeterminate form $\frac{0}{0}$, use rationalization or L'Hôpital's rule. 6. Use L'Hôpital's rule: differentiate numerator and denominator. 7. Derivative numerator: $$2x + 3$$ 8. Derivative denominator: $$2 + \frac{3}{2\sqrt{3x + 4}}$$ 9. Substitute $x = -1$: Numerator: $$2(-1) + 3 = 1$$ Denominator: $$2 + \frac{3}{2\sqrt{3(-1) + 4}} = 2 + \frac{3}{2\sqrt{1}} = 2 + \frac{3}{2} = \frac{7}{2}$$ 10. Limit: $$\frac{1}{\frac{7}{2}} = \frac{2}{7}$$ **Answer:** $$\frac{2}{7}$$ --- **(n) Calculate** $$\lim_{x \to 1} \frac{x - \sqrt{x}}{2 - \sqrt{x + 3}}$$ 2. Substitute $x=1$: Numerator: $$1 - 1 = 0$$ Denominator: $$2 - \sqrt{4} = 2 - 2 = 0$$ 3. Indeterminate form $\frac{0}{0}$, rationalize numerator and denominator. 4. Rationalize numerator: $$x - \sqrt{x} = \sqrt{x}^2 - \sqrt{x} = \sqrt{x}(\sqrt{x} - 1)$$ 5. Rationalize denominator: $$2 - \sqrt{x + 3} = \frac{(2 - \sqrt{x + 3})(2 + \sqrt{x + 3})}{2 + \sqrt{x + 3}} = \frac{4 - (x + 3)}{2 + \sqrt{x + 3}} = \frac{1 - x}{2 + \sqrt{x + 3}}$$ 6. So original limit becomes: $$\lim_{x \to 1} \frac{\sqrt{x}(\sqrt{x} - 1)}{\frac{1 - x}{2 + \sqrt{x + 3}}} = \lim_{x \to 1} \frac{\sqrt{x}(\sqrt{x} - 1)(2 + \sqrt{x + 3})}{1 - x}$$ 7. Note $1 - x = -(x - 1)$ and $\sqrt{x} - 1 = \frac{x - 1}{\sqrt{x} + 1}$ 8. Substitute: $$\frac{\sqrt{x} \cdot \frac{x - 1}{\sqrt{x} + 1} (2 + \sqrt{x + 3})}{1 - x} = \frac{\sqrt{x} (x - 1)(2 + \sqrt{x + 3})}{(\sqrt{x} + 1)(1 - x)}$$ 9. Since $1 - x = -(x - 1)$: $$= \frac{\sqrt{x} (x - 1)(2 + \sqrt{x + 3})}{(\sqrt{x} + 1)(-(x - 1))} = - \frac{\sqrt{x} (2 + \sqrt{x + 3})}{\sqrt{x} + 1}$$ 10. Substitute $x = 1$: $$- \frac{1 (2 + 2)}{1 + 1} = - \frac{4}{2} = -2$$ **Answer:** $$-2$$ --- **(o) Calculate** $$\lim_{x \to 0} \frac{\sin 5x}{\sin 6x}$$ 2. Use limit property $\lim_{x \to 0} \frac{\sin ax}{\sin bx} = \frac{a}{b}$ 3. So: $$\lim_{x \to 0} \frac{\sin 5x}{\sin 6x} = \frac{5}{6}$$ **Answer:** $$\frac{5}{6}$$ --- **(p) Calculate** $$\lim_{x \to 0} \frac{\sin 3x - \sin x}{x}$$ 2. Use sine subtraction formula: $$\sin A - \sin B = 2 \cos \frac{A + B}{2} \sin \frac{A - B}{2}$$ 3. Apply: $$\sin 3x - \sin x = 2 \cos 2x \sin x$$ 4. So limit becomes: $$\lim_{x \to 0} \frac{2 \cos 2x \sin x}{x} = 2 \lim_{x \to 0} \cos 2x \cdot \lim_{x \to 0} \frac{\sin x}{x}$$ 5. Use limits: $$\lim_{x \to 0} \cos 2x = 1, \quad \lim_{x \to 0} \frac{\sin x}{x} = 1$$ 6. So: $$2 \cdot 1 \cdot 1 = 2$$ **Answer:** $$2$$ --- **(q) Calculate** $$\lim_{x \to 4} \frac{x - 4}{(\sqrt{x} - 2)^2}$$ 2. Note that $\sqrt{4} = 2$, so denominator tends to zero. 3. Rewrite denominator: $$(\sqrt{x} - 2)^2 = (\sqrt{x} - 2)(\sqrt{x} - 2)$$ 4. Substitute $t = \sqrt{x}$, so $x = t^2$, and as $x \to 4$, $t \to 2$. 5. Rewrite limit: $$\lim_{t \to 2} \frac{t^2 - 4}{(t - 2)^2} = \lim_{t \to 2} \frac{(t - 2)(t + 2)}{(t - 2)^2}$$ 6. Cancel common factor: $$\frac{\cancel{(t - 2)}(t + 2)}{\cancel{(t - 2)}(t - 2)} = \frac{t + 2}{t - 2}$$ 7. Substitute $t = 2$: Denominator $t - 2 = 0$, so limit diverges. 8. Check left and right limits: - As $t \to 2^+$, denominator $> 0$, numerator $= 4$, so limit $\to +\infty$ - As $t \to 2^-$, denominator $< 0$, numerator $= 4$, so limit $\to -\infty$ **Answer:** Limit does not exist (infinite discontinuity). --- **2. Find asymptotes of curves:** **(a) $$y = \frac{1}{x^2 + 1}$$** - Denominator $x^2 + 1 > 0$ for all real $x$, so no vertical asymptotes. - Horizontal asymptote as $x \to \pm \infty$: $$\lim_{x \to \pm \infty} y = \lim_{x \to \pm \infty} \frac{1}{x^2 + 1} = 0$$ - So horizontal asymptote: $$y = 0$$ **(b) $$y = x - 1 + \frac{1}{x + 1}$$** - Vertical asymptote where denominator zero: $x + 1 = 0 \Rightarrow x = -1$ - Horizontal or oblique asymptote: As $x \to \pm \infty$, $$y \approx x - 1$$ - So oblique asymptote: $$y = x - 1$$ **(c) $$y = 3x + \sqrt{x^2 - 6x}$$** - For large $x$, factor inside root: $$\sqrt{x^2 - 6x} = |x| \sqrt{1 - \frac{6}{x}}$$ - For $x \to +\infty$, $|x| = x$, so: $$y = 3x + x \sqrt{1 - \frac{6}{x}} = x(3 + \sqrt{1 - \frac{6}{x}})$$ - As $x \to +\infty$, $$\sqrt{1 - \frac{6}{x}} \to 1$$ - So: $$y \sim x(3 + 1) = 4x$$ - Oblique asymptote: $$y = 4x$$ - For $x \to -\infty$, $|x| = -x$, so: $$y = 3x - x \sqrt{1 - \frac{6}{x}} = x(3 - \sqrt{1 - \frac{6}{x}})$$ - As $x \to -\infty$, $$\sqrt{1 - \frac{6}{x}} \to 1$$ - So: $$y \sim x(3 - 1) = 2x$$ - Oblique asymptote: $$y = 2x$$ **Answer:** - (a) Horizontal asymptote: $$y = 0$$ - (b) Vertical asymptote: $$x = -1$$, oblique asymptote: $$y = x - 1$$ - (c) Oblique asymptotes: $$y = 4x$$ as $x \to +\infty$, and $$y = 2x$$ as $x \to -\infty$