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de2de:chapter-5 [2012/06/25 11:32] (current)
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 +====Differential Equations to Difference Equations====
  
 +===Answers for selected problems===
 +
 +**Chapter 5**
 +
 +----
 +
 +**Section 5.1**
 +
 +1. $ \displaystyle{\lim_{h \to 0}\frac{\tan^2(h)}{h} = 0}$ 
 +
 +
 +3. $ \displaystyle{\lim_{h \to 0}\frac{h^2\sin(h)}{h^2} = 0}$ and $ \displaystyle{\lim_{h \to 0}\frac{h^2\sin(h)}{h^3} = 1}$
 +
 +
 +5. $ \displaystyle{\lim_{h \to 0}\frac{\sin^2(3h)}{h^2} = 9}$ 
 +
 +
 +7. (a) $ \displaystyle{P_4(x) = 2x - \frac{4x^3}{3}}$ ​
 +
 +%%(c)%% $ \displaystyle{P_4(z) = 2 + \frac{1}{4}(z - 4) - \frac{1}{64}(z - 4)^2 + \frac{1}{512}(z - 4)^3 - \frac{5}{16384}(z - 4)^4}$
 +
 +(e) $ \displaystyle{P_4(x) = -(x - \pi) + \frac{1}{6}(x - \pi)^3}$ ​
 +
 +(g) $ \displaystyle{P_4(x) = 1 - (x - 1) + (x - 1)^2 - (x - 1)^3 + (x - 1)^4}$ ​
 +
 +(h) $ \displaystyle{P_4(x) = -9 + 2x + 3x^2}$ ​
 +
 +(i) $ \displaystyle{P_4(x) = 1 - x^2 + x^4}$ 
 +
 +(k) $ \displaystyle{P_4(t) = 1 - 2(t - 1) + 3(t - 1)^2 - 4(t - 1)^3 + 5 (t - 1)^4}$ ​
 +
 +(m) $ \displaystyle{P_4(t) = 1 + \frac{1}{4}t^2 + \frac{5}{24}t^4}$ ​
 +
 +(n) $ \displaystyle{P_4(x) = -5 - (x - 1)  + 7(x - 1)^2 + 8(x - 1)^3 +3(x - 1)^4}$
 +
 +----
 +
 +**Section 5.2**
 +
 +1. (a) $ \displaystyle{P_5(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!}}$
 +
 +$ \sin(0.8) \approx P_5(0.8) = 0.717397 $, rounded to 6 decimal places
 +
 +$ \displaystyle{|\sin(0.8) - P_5(0.8)| \le \frac{0.8^7}{7!} = 0.000041610}$,​ rounded to 9 decimal places ​
 +
 +%%(c)%% $ \displaystyle{P_5(x) = 2x - \frac{4}{3}x^3 + \frac{4}{15}x^5}$
 +
 +$ \sin(-1) \approx P_5(-0.5) = -0.841667 $, rounded to 6 decimal places
 +
 +$ \displaystyle{|\sin(-1) - P_5(-0.5)| \le \frac{2^7(0.5)^7}{7!} = 0.000198413}$,​ rounded to 9 decimal places ​
 +
 +(e) $ \displaystyle{P_5(x) ​ = 3 + \frac{1}{6}(x - 9) - \frac{1}{216}(x - 9)^2 + \frac{1}{3888}(x - 9)^3}$
 +
 +$ \displaystyle{- \frac{5}{279,​936}(x - 9)^4 + \frac{7}{5,​038,​848}(x - 9)^5}$
 +
 +$ \sqrt{10} \approx P_5(10) = 3.1622778 $, rounded to 7 decimal places
 +
 +$ \displaystyle{|\sqrt{10} - P_5(10)| \le \frac{945}{(64)(3^{11})(6!)} = 0.000000115767}$,​ rounded to 12 decimal places ​
 +
 +(g) $ \displaystyle{P_5(x) =  1 - (x - 1) + (x - 1)^2 - (x - 1)^3 + (x - 1)^4 - (x - 1)^5}$
 +
 +$ \displaystyle{\frac{1}{0.8} \approx P_5(0.8) = 1.24992}$, rounded to 5 decimal places
 +
 +$ \displaystyle{\left|\frac{1}{0.8} - P_5(0.8)\right| \le \frac{720(0.2)^6}{(0.8)^7(61)} = 0.000305176}$,​ rounded to 9 decimal places ​
 +
 +
 +3. $ \sin(-1.3) \approx P_9(-1.3) = -0.963559 $, rounded to 6 decimal places ​
 +
 +
 +5. $ \displaystyle{P_9(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \frac{x^9}{9!}}$ ​
 +
 +
 +7. (a) $ \displaystyle{P_{10}(x) = 1 + x + \frac{x^2}{2} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \frac{x^6}{6!} + \frac{x^7}{7!} + \frac{x^8}{8!} + \frac{x^9}{9!} + \frac{x^{10}}{10!}}$ ​
 +
 +(b) $ E(1) \approx P_{10}(1) = 2.7182818011 $, rounded to 10 decimal places ​
 +
 +%%(c)%% $ \displaystyle{|E(1) - P_{10}(1)| \le \frac{3}{11!} = 0.00000007515633}$,​ rounded to 14 decimal places ​
 +
 +(d) $ \displaystyle{P_{11}(x) = 1 + x + \frac{x^2}{2} \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \frac{x^6}{6!} + \frac{x^7}{7!} + \frac{x^8}{8!} + \frac{x^9}{9!} + \frac{x^{10}}{10!} + \frac{x^{11}}{11!}}$ ​
 +
 +
 +8. $ \displaystyle{\int_0^3 \frac{\sin(x)}{x} \ dx \approx \int_0^3 \frac{P_9(x)}{x} \ dx = \int_0^3 \left(1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \frac{x^6}{7!} + \frac{x^8}{9!}\right)dx = 1.849037}$, rounded to 6 decimal places ​
 +
 +
 +9. (a) $ \displaystyle{P_6(x) = x^2 - \frac{x^6}{6}}$ ​
 +
 +(b) $ \displaystyle{P_7(x) = \frac{x^3}{3} - \frac{x^7}{42}}$
 +
 +----
 +
 +**Section 5.3**
 +
 +1. (a) The series converges because it is a multiple of a geometric series with ratio $ r = \dfrac{1}{5} $. Moreover, $ \displaystyle{\sum_{n=0}^\infty \frac{3}{5^n} = \frac{15}{4}}$ ​
 +
 +%%(c)%% The series converges because it is the sum of two geometric series with ratios $ r =\dfrac{1}{2} $ for the first and $ r = \dfrac{1}{3} $ for the second. Moreover, $ \displaystyle{\sum_{n=1}^\infty \left(\frac{5}{2^n} + \frac{2}{3^n}\right) = 6}$ 
 +
 +(e) Since $ \displaystyle{\lim_{n \to \infty}\left(1 - \frac{1}{n}\right) = 1}$, the series diverges by the $ n $th term test for divergence. ​
 +
 +(g) Since $ \dfrac{(-1)^n}{1000} $ does not have a limit as $ n \rightarrow \infty $, the series diverges by the $ n $th term test for divergence. ​
 +
 +
 +2. (a) The series converges because it is a multiple of a $ p $-series with $ p = 3 $. 
 +
 +%%(c)%% The series converges because it is the sum of multiples of two $ p $-series with $ p = 2 $ for the first and $ p = 4 $ for the second. ​
 +
 +(e) The series diverges because it is a $ p $-series with $ p = \dfrac{1}{3} $. 
 +
 +(g) The series diverges because it is a multiple of a $ p $-series with $ p = \dfrac{1}{2}. $ 
 +
 +
 +5. (a) The series converges.
 +
 +%%(c)%% The integral diverges.
 +
 +----
 +
 +**Section 5.4**
 +
 +1. (a) The series converges because it is a multiple of a geometric series with ratio $ r = \dfrac{1}{2} $. 
 +
 +%%(c)%% The series diverges by comparion with a $ p $-series with $ p = \dfrac{1}{2} $. 
 +
 +(e) The series converges because it is a geometric series with ratio $ r = \dfrac{1}{\pi} $. 
 +
 +(g) The series converges because is the sum of a convergent geometric series (with ratio $ r = \dfrac{1}{3} $) and a convergent $ p $-series (with $ p = \dfrac{3}{2} $). 
 +
 +
 +2. (a) The series diverges by comparison with the harmonic series. ​
 +
 +%%(c)%% The series converges by the limit comparison test with a $ p $-series with $ p = 3 $. 
 +
 +(e) The series converges by the limit comparison test with a geometric series with ratio $ r = \dfrac{1}{3} $. 
 +
 +(g) The series diverges by the limit comparison test with a $ p $-series with $ p = 2 $. 
 +
 +
 +4. The series $ \displaystyle{\sum_{n=1}^\infty \frac{1}{4n^5 - 2}}$ converges by the limit comparison test with a $ p $-series with $ p = 5 $. Hence $ \displaystyle{\int_1^\infty \frac{1}{4x^5 - 2} \ dx}$ convergess by the integral test.
 +
 +----
 +
 +**Section 5.5**
 +
 +1. (a) The series converges by the ratio test: $ \rho = \dfrac{1}{2} $. 
 +
 +%%(c)%% The series converges by the ratio test: $ \rho = 0 $. 
 +
 +(e) The series converges by the ratio test: $ \rho = 0 $. 
 +
 +(g) The series converges by the ratio test: $ \rho = 0 $. 
 +
 +(h) The series diverges by the ratio test: $ \rho = \pi $. 
 +
 +
 +2. (a) The series diverges by the limit comparison test with a $ p $-series with $ p = \dfrac{1}{2} $.
 +
 +%%(c)%% The series converges by the ratio test: $ \rho = 0 $. 
 +
 +(e) The series converges by the ratio test: $ \rho = \dfrac{1}{4} $.
 +
 +(g) The series diverges by the $ n $th term test for divergence since $\displaystyle{\lim_{n \to \infty}\frac{3n + 1}{2n - 1} = \frac{3}{2}}$.
 +
 +
 +3. (a) $ (-1, 1) $ 
 +
 +
 +4. (a) $ (-\infty, \infty) $ 
 +
 +
 +5. The series converges by the ratio test since $\displaystyle{\rho = \lim_{n \to \infty}\frac{a_{n+1}}{a_n} = \frac{1}{2}}$.
 +
 +----
 +
 +**Section 5.6**
 +
 +1. (a) The series converges absolutely ($ p $-series with $ p = 3 $), and so converges. ​
 +
 +%%(c)%% The series diverges by the $ n $th term for divergence since
 +<table align="​center">​
 +<​tr><​td align="​center">​$ \displaystyle{\lim_{n \to \infty}\frac{3n^2 - 1}{4n^2 + 2} = \frac{3}{4}}$.</​td></​tr>​
 +</​table> ​
 +
 +(e) The series does not converge absolutely (limit comparison test with a $ p $-series with $ p = \dfrac{1}{2} $), but does converge conditionally by Leibniz'​s theorem. Hence the series converges. ​
 +
 +(g) Since $ \displaystyle{\lim_{n \to \infty}\frac{n!}{2^n} = \infty}$, the terms $ \dfrac{(-1)^n n!}{2^n} $ do not have a limit as $ n \rightarrow \infty $. Hence the series diverges by the $ n $th term test for divergence. ​
 +
 +
 +2. (a) The series converges absolutely (limit comparison with a $ p $-series with $ p = 3 $), and so converges. ​
 +
 +%%(c)%% The series converges absolutely (ratio test, $ \rho = 0 $), and so converges. ​
 +
 +(e) The series converges absolutely (ratio test, $ \rho = 0 $), and so converges. ​
 +
 +(g) Since $ \displaystyle{\lim_{n \to \infty}\frac{n + 1}{2n - 1} = \frac{1}{2}}$,​ the terms $ \dfrac{(-1)^n (n + 1)}{2n - 1} $ do not have a limit as $ n \rightarrow \infty $. Hence the series diverges by the the $ n $th term test for divergence. ​
 +
 +
 +3. (a) $ \displaystyle{s_{15} = \frac{34,​361,​893,​981}{93,​405,​312,​000} =  0.367879441171397}$,​ to 15 decimal places ​
 +
 + (b) $ |s - s_{15}| \le \dfrac{1}{16!} = 4.779 \times 10^{-14} $, rounded to 17 decimal places ​
 +
 +%%(c)%% $ s_9 = \dfrac{16,​687}{45,​360} = 0.367879 $ to 6 decimal places
 +
 +----
 +
 +**Section 5.7**
 +
 +1. (a) The interval of convergence is $ (-1, 1) $.
 +
 +The series converges at $ x = -1 $ and diverges at $ x = 1 $.
 +
 +The series converges absolutely on $ (-1, 1) $, converges conditionally at $ x = -1 $, and diverges for $ x < -1 $ and $ x \ge 1 $.
 +
 +$ \displaystyle{\sum_{n = 1}^\infty \frac{x^n}{n} = x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + \frac{x^5}{5} + \cdots}$ ​
 +
 +%%(c)%% The interval of convergence is $ (-1, 1) $.
 +
 +The series diverges at both $ x = -1 $ and $ x = 1 $.
 +
 +The series converges absolutely on $ (-1, 1) $ and diverges for $ x \le -1 $ and $ x \ge 1 $.
 +
 +$ \displaystyle{\sum_{n = 1}^\infty nx^n = x + 2x^2 + 3x^3 + 4x^4 + 5x^5 + \cdots}$ ​
 +
 +(e) The interval of convergence is $ (1, 3) $.
 +
 +The series converges at $ x = 1 $ and diverges at $ x = 3 $.
 +
 +The series converges absolutely on $ (1, 3) $, converges conditionally at $ x = 1 $, and diverges for $ x < 1 $ and $ x \ge 3 $.
 +
 +$ \displaystyle{\sum_{n = 0}^\infty \frac{(x-2)^{n+1}}{n+1} = (x - 2) + \frac{(x-2)^2}{2} + \frac{(x-2)^3}{3} + \frac{(x-2)^4}{4} + \frac{(x-2)^5}{5} + \cdots}$ ​
 +
 +(g) The interval of convergence is $ (0, 2) $.
 +
 +The series converges at both $ x = 0 $ and $ x = 2 $.
 +
 +The series converges absolutely on $ [0, 2] $ and diverges for $ x < 0 $ and $ x > 2 $.
 +
 +$ \displaystyle{\sum_{n = 1}^\infty \frac{(x-1)^n}{n^2} = (x-1) + \frac{(x-1)^2}{4} + \frac{(x-1)^3}{9} + \frac{(x-1)^4}{16} + \frac{(x-1)^5}{25} + \cdots}$ ​
 +
 +
 +2. (a) The interval of convergence is $ (-\infty, \infty) $.
 +
 +The series converges absolutely on $ (-\infty, \infty) $.
 +
 +$ \displaystyle{\sum_{n = 0}^\infty \frac{(x-3)^n}{n!} = 1 + (x - 3) + \frac{(x-3)^2}{2} + \frac{(x-3)^3}{6} + \frac{(x-3)^4}{24} + \cdots}$ ​
 +
 +%%(c)%% The interval of convergence is $ (-\sqrt{2}, \sqrt{2}) $.
 +
 +The series diverges at both $ x = -\sqrt{2} $ and $ x = \sqrt{2} $.
 +
 +The series converges absolutely on $ (-\sqrt{2}, \sqrt{2}) $ and diverges for $ x \le -\sqrt{2} $ and $ x \ge \sqrt{2} $.
 +
 +$ \displaystyle{\sum_{n = 0}^\infty \frac{x^{2n}}{2^n} = 1 + \frac{x^2}{2} + \frac{x^4}{4} + \frac{x^6}{8} + \frac{x^8}{16}+ \cdots}$ ​
 +
 +(e) The interval of convergence is $ (-1, 1) $.
 +
 +The series diverges at both $ x = -1 $ and $ x = 1 $.
 +
 +The series converges absolutely on $ (-1, 1) $ and diverges for $ x \le -1 $ and $ x \ge 1 $.
 +
 +$ \displaystyle{\sum_{n = 1}^\infty \frac{x^{2n+1}}{n} = x^3 + \frac{x^5}{2} + \frac{x^7}{3} + \frac{x^9}{4} + \frac{x^{11}}{5} + \cdots}$ ​
 +
 +(g) The interval of convergence is $ \displaystyle{\left(-\frac{1}{3},​ \frac{1}{3}\right)}$.
 +
 +The series diverges at both $ x = -\dfrac{1}{3} $ and $ x = \dfrac{1}{3} $.
 +
 +The series converges absolutely on $ \displaystyle{\left(-\frac{1}{3},​ \frac{1}{3}\right)}$ and diverges for $ x \le -\dfrac{1}{3} $ and $ x \ge \dfrac{1}{3} $.
 +
 +$ \displaystyle{\sum_{n = 1}^\infty 3^nx^n = 3x + 9x^2 + 27x^3 + 81x^4 + 243x^5 + \cdots}$ ​
 +
 +
 +3. (a) For $ -1 < x < 1 $,
 +\[
 +\frac{1}{1 + x} = \sum_{n=0}^\infty (-1)^nx^n = 1 - x + x^2 - x^3 + x^4 - \cdots.
 +\]
 +
 +(b) $ f^{(35)}(0) = -35! $ 
 +
 +%%(c)%% $ \displaystyle{\int_0^x \frac{1}{1 + t} \ dt = \sum_{n=0}^\infty \frac{(-1)^n x^{n+1}}{n + 1} = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5} - \cdots}$
 +
 +The series converges absolutely on $ (-1, 1) $, converges conditionally at $ x = 1 $, and diverges for $ x \le -1 $ and $ x > 1 $. 
 +
 +(d) $ \displaystyle{\int_0^1 \frac{1}{1 + t} \ dt = \sum_{n=0}^\infty \frac{(-1)^n}{n + 1} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \cdots}$
 +
 +$ \displaystyle{\int_0^1 \frac{1}{1 + t} \ dt \approx \sum_{n=0}^{998} \frac{(-1)^n}{n + 1} = 0.6936}$, rounded to 4 decimal places ​
 +
 +
 +5. $ \displaystyle{\cos(1) = \sum_{n=0}^\infty \frac{(-1)^{2n}}{(2n)!} = 1 - \frac{1}{2} + \frac{1}{4!} - \frac{1}{6!} + \frac{1}{8!} - \cdots}$
 +
 +$ \displaystyle{\cos(1) \approx 1 - \frac{1}{2} + \frac{1}{24} - \frac{1}{720} + \frac{1}{40,​320} = 0.5403026}$,​ rounded to 7 decimal places ​
 +
 +
 +7. $ \displaystyle{\sum_{n=1}^\infty \frac{n}{2^{n-1}} = \frac{1}{\left(1 - \frac{1}{2}\right)^2} = 4}$ 
 +
 +
 +8. (a) $ \displaystyle{A = \sum_{n=1}^\infty \frac{n}{2^n} = \frac{1}{2}\sum_{n=1}^\infty \frac{n}{2^{n-1}} = 2}$ 
 +
 +(b) $ \displaystyle{A = q\sum_{n=1}^\infty np^{n-1} = \frac{q}{(1 - p)^2} = \frac{1}{q}}$ ​
 +
 +----
 +
 +**Section 5.8**
 +
 +2. (a) For $ -\infty < x < \infty $,
 +\[
 +\cos(x^2) = \sum_{n=0}^\infty \frac{(-1)^nx^{4n}}{(2n)!} = 1 - \frac{x^4}{2} + \frac{x^8}{4!} - \frac{x^{12}}{6!} + \frac{x^{16}}{8!} - \cdots .
 +\]
 +
 +%%(c)%% For $ -1 < t < 1 $,
 +\[
 +\frac{1}{1 - t^2} = \sum_{n=0}^\infty t^{2n} = 1 + t^2 + t^4 + t^6 + t^8 + \cdots .
 +\]
 +
 +(d) For $ -\dfrac{1}{2} < x < \dfrac{1}{2} $,
 +\[
 +\frac{1}{2x - 1} = -\sum_{n=1}^\infty 2^nx^n = -1 - 2x - 4x^2 - 8x^3 - 16x^4 - \cdots .
 +\]
 +
 +(e) For $ -1 < t < 1 $,
 +\[
 +\frac{1}{(1 + t)^2} = \sum_{n=1}^\infty (-1)^{n+1}nt^{n-1} = 1 - 2t + 3t^2 - 4t^3 + 5t^4 - \cdots .
 +\]
 +
 +(g) For $ -\infty < x < \infty $,
 +\[
 +f(x) = \sum_{n=1}^\infty \frac{(-1)^{n+1}x^{2n-1}}{(2n)!} = \frac{x}{2} - \frac{x^3}{4!} + \frac{x^5}{6!} - \frac{x^7}{8!} + \frac{x^9}{10!} - \cdots .
 +\]
 +
 +
 +3. (a) For $ -\infty < x < \infty $,
 +\[
 +\cos^2(x) = 1 + \sum_{n=1}^\infty \frac{(-1)^n2^{2n-1}x^{2n}}{(2n)!} = 1 - x^2 + \frac{x^4}{3} - \frac{2x^6}{45} + \frac{x^8}{315} - \cdots .
 +\]
 +
 +(b) $ \displaystyle{P_8(x) = 1 - x^2 + \frac{x^4}{3} - \frac{2x^6}{45} + \frac{x^8}{315}}$ ​
 +
 +
 +5. (a) For $ -\infty < x < \infty $,
 +\[
 +\sin(x^2) = \sum_{n=0}^\infty \frac{(-1)^nx^{4n+2}}{(2n+1)!} = x^2 - \frac{x^6}{3!} + \frac{x^{10}}{5!} - \frac{x^{14}}{7!} + \frac{x^{18}}{9!} - \cdots .
 +\]
 +
 +(b) $ \displaystyle{P_{10}(x) = x^2 - \frac{x^6}{3!} + \frac{x^{10}}{5!}}$ ​
 +
 +%%(c)%% For $ -\infty < x < \infty $,
 +\[
 +S(x) = \sum_{n=0}^\infty \frac{(-1)^nx^{4n+3}}{(4n+3)(2n+1)!} = \frac{x^3}{3} - \frac{x^7}{7(3!)} + \frac{x^{11}}{11(5!)} - \frac{x^{15}}{15(7!)} + \frac{x^{19}}{19(9!)} - \cdots .
 +\]
 +
 +(d) $ \displaystyle{P_{11}(x) = \frac{x^3}{3} - \frac{x^7}{7(3!)} + \frac{x^{11}}{11(5!)}}$ ​
 +
 +(e) $ S(1) \approx P_{15}(1) = 0.310268 $, rounded to 6 deciimal places ​
 +
 +
 +7. $ \displaystyle{\frac{d^9}{dx^9}\mathrm{Si}(x)\Bigg|_{x=0} = \frac{1}{9}}$
 +
 +----
 +
 +**Section 5.9**
 +
 +1. (a) $ \displaystyle{\lim_{x \to 0} \frac{\sin(3x)}{x} = 3}$ 
 +
 +%%(c)%% $ \displaystyle{\lim_{x \to 0} \frac{\cos(x) - 1 + \dfrac{x^2}{2}}{x^4} = \frac{1}{24}}$ ​
 +
 +(e) $ \displaystyle{\lim_{u \to 0} \frac{\tan(u)}{\sin(u)} = 1}$ 
 +
 +(g) $ \displaystyle{\lim_{y \to 0} \frac{\tan(3y)}{\tan(5y)} = \frac{3}{5}}$ ​
 +
 +(i) $ \displaystyle{\lim_{x \to 0} \frac{\sqrt{1+x} - 1}{3x} = \frac{1}{6}}$ ​
 +
 +
 +2. (a) $ \displaystyle{\lim_{x \to 0} \frac{\sin(5x)}{3x} = \frac{5}{3}}$ ​
 +
 +%%(c)%% $ \displaystyle{\lim_{x \to 0} \frac{1 - \sec(x)}{x} = 0}$ 
 +
 +(e) $ \displaystyle{\lim_{x \to 0^+} \frac{\sin(2x)}{\sqrt{x}} = 0}$ 
 +
 +(g) $ \displaystyle{\lim_{x \to \infty}x\sin\left(\frac{1}{x^2}\right) = 0}$ 
 +
 +
 +3.(a) $ \displaystyle{\lim_{x \to \infty} \frac{3x^2 - 2x + 1}{16x^2 + 2} = \frac{3}{16}}$ ​
 +
 +%%(c)%% $ \displaystyle{\lim_{x \to 0}\frac{1 - \dfrac{\sin(x)}{x}}{3x^2} = \frac{1}{18}}$ ​
 +
 +(e) $ \displaystyle{\lim_{x \to 0} \frac{(1+x)^{\frac{1}{3}} - 1 - \dfrac{x}{3}}{x^2} = -\frac{1}{9}}$ ​
 +
 +(g) $ \displaystyle{\lim_{x \to 1}\frac{\sqrt{x} - 1}{x - 1} = \frac{1}{2}}$ ​
 +
 +(i) $ \displaystyle{\lim_{t \to 0}\frac{\cos(t) - 1 + \dfrac{t^2}{2}}{t^2\sin(t^2)} = \frac{1}{24}}$
 +
 +----