Answer to Question #343658 in Real Analysis for Mr Alex

Question

a) Find $\ lim_{x\to 0}\frac{\left(tanxsec^2x-x\right)}{x^3}$

 

 b) Examine whether the equation, x³- 11x +9 =0 has a real root in the interval [0,1]

 c) Check whether the following series are convergent or not (4) 

 i) $\sum _{n=1}^{\infty }\:\frac{\left(3n-1\right)}{7^n}$

(ii) $\sum _{n=1}^{\infty }\frac{\left(\:\sqrt{n^2+3}-\sqrt{n^2-3}\right)}{\sqrt{n}}\:$

Answer 1

a)

$$\lim\limits_{x\to 0}\dfrac{\tan x\sec^2 x-x}{x^3}=\lim\limits_{x\to 0}\dfrac{\tan x(1)^2-x}{x^3}$$

$$=\lim\limits_{x\to 0}\dfrac{x+\dfrac{x^3}{3}+\dfrac{2x^5}{15}+...-x}{x^3}$$

$$=\lim\limits_{x\to 0}\dfrac{\dfrac{x^3}{3}+\dfrac{2x^5}{15}+...}{x^3}$$

$$=\lim\limits_{x\to 0}(\dfrac{1}{3}+\dfrac{2x^2}{15}+...)=\dfrac{1}{3}$$

b)

$f(x)=x^3-11x+9$ is continuous on $[0, 1].$

$$f(0)=(0)^3-11(0)+9=9>0$$

$$f(1)=(1)^3-11(1)+9=-1<0$$

Then by the Intermediate Value Theorem there exists a number $c$ in $(0, 1)$ such that $f(c)=0.$

Therefore the equation, $x^3- 11x +9 =0$ has a real root in the interval $[0,1]$ by the Intermediate Value Theorem.

c)

(i)

$$\displaystyle\sum_{i=1}^{\infin}\dfrac{3n-1}{7^n}$$

Use the Ratio Test

$$|\dfrac{a_{n+1}}{a_n}|=|\dfrac{\dfrac{3(n+1)-1}{7^{n+1}}}{\dfrac{3n-1}{7^n}}|=\dfrac{1}{7}(\dfrac{3n+2}{3n-1})\to\dfrac{1}{7}<1$$

Thus, by the Ratio Test, the given series is absolutely convergent and therefore convergent.

(ii)

$$\displaystyle\sum_{i=1}^{\infin}\dfrac{\sqrt{n^2+3}-\sqrt{n^2-3}}{\sqrt{n}}$$

$$=\displaystyle\sum_{i=1}^{\infin}\dfrac{n^2+3-(n^2-3)}{\sqrt{n}(\sqrt{n^2+3}+\sqrt{n^2-3})}$$

$$=\displaystyle\sum_{i=1}^{\infin}\dfrac{6}{\sqrt{n}(\sqrt{n^2+3}+\sqrt{n^2-3})}$$

Use the Limit Comparison Test

Take $a_n=\dfrac{6}{\sqrt{n}(\sqrt{n^2+3}+\sqrt{n^2-3})}, b_n=\dfrac{1}{n}$

$$\lim\limits_{n\to \infin}\dfrac{a_n}{b_n}=\lim\limits_{n\to \infin}\dfrac{\dfrac{6}{\sqrt{n}(\sqrt{n^2+3}+\sqrt{n^2-3})}}{\dfrac{1}{n}}$$

$$=\lim\limits_{n\to \infin}\dfrac{6}{\sqrt{1+3/n^2}+\sqrt{1-3/n^2}}=3$$

The harmonic series $\displaystyle\sum_{i=1}^{\infin}\dfrac{1}{n}$ is divergent.

Therefore the series $\displaystyle\sum_{i=1}^{\infin}\dfrac{\sqrt{n^2+3}-\sqrt{n^2-3}}{\sqrt{n}}$ is divergent by the Limit Comparison Test.