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# Answer to Question #343658 in Real Analysis for Mr Alex

Question #343658

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}}\\:"

1
2022-05-24T15:56:33-0400

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.

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