Exercises

Prove the following:

1. $\lim{x \to \infty} \left ( \frac{x + 1}{x} \right ) = 1$.

   Solution:

   $$\begin{array}{ll} \lim_{x \to \infty} & = \lim_{x \to \infty... ...y} \left ( \frac{1}{x} \right ) \\ & = 1 + 0 = 1, \end{array}$$
   by Theorem 3.8.1

2. $\lim_{x \to \infty} \left ( \frac{x^2 + 2x}{5 - 3x^2} \right ) = -\frac{1}{3}$.

   Solution:

   $$\lim_{x \to \infty} \left ( \frac{x^2 + 2x}{5 - 3x^2} \right ) = \lim_{x \to \infty} \left ( \frac{1 + \frac{2}{x}}{\frac{5}{x^2} - 3} \right )$$
   [ Dividing both numerator and denominator by $x^2$.]
   $$= \frac{\lim_{x \to \infty} \left ( 1 + \frac{2}{x} \right )}{\lim_{x \to \infty} \left ( \frac{5}{x^2} - 3 \right )}$$
   by Theorem 3.8.3
   $$= \frac{\lim_{x \to \infty} (1) + \lim_{x \to \infty} \left ( \fr... ...^2} \right ) - \lim_{x \to \infty} (3)} = \frac{1 + 0}{0 - 3} = -\frac{1}{3},$$
   by Theorem 3.8.1.

3. $\lim_{x \to 1} \frac{x^2 - 2x + 5}{x^2 + 7} = \frac{1}{2}$.

4. $\lim_{x \to 0} \frac{3x^3 + 6x^2}{2x^4 - 15x^2} = -\frac{2}{5}$.

5. $\lim_{x \to -2} \frac{x^2 + 1}{x + 3} = 5$.

6. $\lim_{h \to 0} (3ax^2 - 2hx + 5h^2) = 3ax^2$.

7. $\lim_{x \to \infty} (ax^2 + bx + c) = \infty$.

8. $\lim_{k \to 0} \frac{(x - k)^2 - 2kx^3}{x(x + k)} = 1$.

9. $\lim_{x \to \infty} \frac{x^2 + 1}{3x^2 + 2x - 1} = \frac{1}{3}$.

10. $\lim_{x \to \infty} \frac{3 + 2x}{x^2 - 5x} = 0$.

11. $\lim_{\alpha \to \frac{\pi}{2}} \frac{\cos(\alpha - a)}{\cos(2\alpha - a)} = -\tan \alpha$.

12. $\lim_{x \to \infty} \frac{ax^2 + bx + c}{dx^2 + ex + f} = \frac{a}{d}$.

13. $\lim_{z \to 0} \frac{a}{2} (e^{\frac{z}{a}} + e^{-\frac{z}{a}}) = a$.

14. $\lim_{x \to 0} \frac{2x^3 + 3x^2}{x^3} = \infty$.

15. $\lim_{x \to \infty} \frac{5x^2 - 2x}{x} = \infty$.

16. $\lim_{y \to \infty} \frac{y}{y + 1} = 1$.

17. $\lim_{n \to \infty} \frac{n(n + 1)}{(n + 2)(n + 3)} = 1$.

18. $\lim_{s \to 1} \frac{s^3 - 1}{s - 1} = 3$.

19. $\lim_{h \to 0} \frac{(x + h)^n - x^n}{h} = nx^{n-1}$.

20. $\lim_{h = 0} \left [ \cos(\theta + h) \frac{\sin h}{h} \right ] = \cos \theta$.

21. $\lim_{x \to \infty} \frac{4x^2 - x}{4 - 3x^2} = -\frac{4}{3}$.

22. $\lim_{\theta \to 0} \frac{1 - \cos \theta}{\theta^2} = \frac{1}{2}$.

23. $\lim_{x \to a} \frac{1}{x - a} = -\infty$, if $x$ is increasing as it approaches the value $a$.

24. $\lim_{x \to a} \frac{1}{x - a} = +\infty$, if $x$ is decreasing as it approaches the value $a$.

Here is an example of the limit in Exercise 22 using SAGE:

[fontsize=\scriptsize,fontfamily=courier,fontshape=tt,frame=single,label=\sage]

sage: theta = var("theta")
sage: limit((1 - cos(theta))/(theta^2),theta=0)
1/2

In other words, for small values of $\theta$, $\cos(\theta) \cong 1+\frac{1}{2}\theta^2.$