Solutions for Finding Limits: Numerical and Graphical Approaches
Solutions to Try Its
1. [latex]a=5[/latex], [latex]f\left(x\right)=2{x}^{2}-4[/latex], and [latex]L=46[/latex]. 2. a. 0; b. 2; c. does not exist; d. [latex]-2[/latex]; e. 0; f. does not exist; g. 4; h. 4; i. 4 3. [latex]\underset{x\to 0}{\mathrm{lim}}\left(\frac{20\sin \left(x\right)}{4x}\right)=5[/latex]
4. does not exist
Solutions to Odd-Numbered Exercises
1. The value of the function, the output, at [latex]x=a[/latex] is [latex]f\left(a\right)[/latex]. When the [latex]\underset{x\to a}{\mathrm{lim}}f\left(x\right)[/latex] is taken, the values of [latex]x[/latex] get infinitely close to [latex]a[/latex] but never equal [latex]a[/latex]. As the values of [latex]x[/latex] approach [latex]a[/latex] from the left and right, the limit is the value that the function is approaching. 3. –4 5. –4 7. 2 9. does not exist 11. 4 13. does not exist 15. Answers will vary. 17. Answers will vary. 19. Answers will vary. 21. Answers will vary. 23. 7.38906 25. 54.59815 27. [latex]{e}^{6}\approx 403.428794[/latex], [latex]{e}^{7}\approx 1096.633158[/latex], [latex]{e}^{n}[/latex] 29. [latex]\underset{x\to -2}{\mathrm{lim}}f\left(x\right)=1[/latex] 31. [latex]\underset{x\to 3}{\mathrm{lim}}\left(\frac{{x}^{2}-x - 6}{{x}^{2}-9}\right)=\frac{5}{6}\approx 0.83[/latex] 33. [latex]\underset{x\to 1}{\mathrm{lim}}\left(\frac{{x}^{2}-1}{{x}^{2}-3x+2}\right)=-2.00[/latex] 35. [latex]\underset{x\to 1}{\mathrm{lim}}\left(\frac{10 - 10{x}^{2}}{{x}^{2}-3x+2}\right)=20.00[/latex] 37. [latex]\underset{x\to \frac{-1}{2}}{\mathrm{lim}}\left(\frac{x}{4{x}^{2}+4x+1}\right)[/latex] does not exist. Function values decrease without bound as [latex]x[/latex] approaches –0.5 from either left or right. 39. [latex]\underset{x\to 0}{\mathrm{lim}}\frac{7\tan x}{3x}=\frac{7}{3}[/latex]
41. [latex]\underset{x\to 0}{\mathrm{lim}}\frac{2\sin x}{4\tan x}=\frac{1}{2}[/latex]
43. [latex]\underset{x\to 0}{\mathrm{lim}}{e}^{{e}^{-\text{ }\frac{1}{{x}^{2}}}}=1.0[/latex]
45. [latex]\underset{x\to -{1}^{-}}{\mathrm{lim}}\frac{|x+1|}{x+1}=\frac{-\left(x+1\right)}{\left(x+1\right)}=-1[/latex] and [latex]\underset{x\to -{1}^{+}}{\mathrm{lim}}\frac{|x+1|}{x+1}=\frac{\left(x+1\right)}{\left(x+1\right)}=1[/latex]; since the right-hand limit does not equal the left-hand limit, [latex]\underset{x\to -1}{\mathrm{lim}}\frac{|x+1|}{x+1}[/latex] does not exist.
47. [latex]\underset{x\to -1}{\mathrm{lim}}\frac{1}{{\left(x+1\right)}^{2}}[/latex] does not exist. The function increases without bound as [latex]x[/latex] approaches [latex]-1[/latex] from either side.
49. [latex]\underset{x\to 0}{\mathrm{lim}}\frac{5}{1-{e}^{\frac{2}{x}}}[/latex] does not exist. Function values approach 5 from the left and approach 0 from the right.
51. Through examination of the postulates and an understanding of relativistic physics, as [latex]v\to c[/latex], [latex]m\to \infty [/latex]. Take this one step further to the solution,
[latex]\underset{v\to {c}^{-}}{\mathrm{lim}}m=\underset{v\to {c}^{-}}{\mathrm{lim}}\frac{{m}_{o}}{\sqrt{1-\left({v}^{2}/{c}^{2}\right)}}=\infty [/latex]
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