Best solutions to - "Limit for (1−cos⁡x)(k+x)x when x→0"

aanvarendbq28

Answered question

2022-03-23

Limit for

\frac{{(1 - \cos x)}^{(k + x)}}{x}

when

x \to 0

German Ferguson

Beginner2022-03-24Added 18 answers

Hint:

\frac{{(1 - \cos x)}^{(\frac{32}{π} + x)}}{x} = {(1 - \cos x)}^{(\frac{32}{π} + x - 1)} (\frac{1 - \cos x}{x})

Nyla Trujillo

Beginner2022-03-25Added 9 answers

Since

{(1 - \cos x)}^{x} \to 1

x \to 0

(prove this!) the desired limit is equal to the limit of

\frac{{(1 - \cos x)}^{k}}{x}

x \to 0

. Further note that

\frac{1 - \cos x}{x^{2}} \to \frac{1}{2}

hence the desired limit is equal to the limit of

2^{- k} x^{2 k - 1}

. Thus the desired limit is equal to

\frac{1}{\sqrt{2}}

if k=1/2 and it is 0 if k>1/2 and diverges if k<1/2. All this is valid for

x \to 0^{+}

. When we take into account

x \to 0^{-}

then the limit is

- \frac{1}{\sqrt{2}}

for k=1/2.
Thus to conclude, the limit does not exist if

k \leq \frac{1}{2}

and is 0 if k>1/2

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