I should prove using the limit definition thatlimx→0x13cos⁡(1x)=0

Question

anita1415snck · Accepted Answer

You can solve this problem with the Squeeze Theorem.
First, notice that

- 1 \leq \cos (\frac{1}{x}) \leq 1

(the cosine graph never goes beyond these bounds, no matter what you put inside as the argument).
Multiplying through by

x^{\frac{1}{3}}

, we get

- x^{\frac{1}{3}} \leq x^{\frac{1}{3}} \cos (\frac{1}{x}) \leq x^{\frac{1}{3}}

Now, the Squeeze Theorem says

lim_{x \to 0} (- x^{\frac{1}{3}}) \leq lim_{x \to 0}, x^{\frac{1}{3}} \cos (\frac{1}{x}) \leq lim_{x \to 0}, x^{\frac{1}{3}},

so we investigate the left- and right-most limits.
Since,

x^{\frac{1}{3}}

is continuous on

[0, \infty)

,

lim_{x \to 0}, x^{\frac{1}{3}} = lim_{x \to 0} (- x^{\frac{1}{3}}) = 0

Finally, we have

0 \leq lim_{x \to 0}, x^{\frac{1}{3}} \cos (\frac{1}{x}) \leq 0

which forces us to conclude

lim_{x \to 0}, x^{\frac{1}{3}} \cos (\frac{1}{x}) = 0

I should prove using the limit definition that