Guys help(｡ŏ﹏ŏ)

Presumably, you mean, 'prove this'.
[MATH]\sin^2 \left(\frac{1}{x} \right)≤1[/MATH] so [MATH]x^2\sin^2 \left(\frac{1}{x} \right)≤x^2[/MATH] and the question becomes trivial.
 
lex said:
Presumably, you mean, 'prove this'.
[MATH]\sin^2 \left(\frac{1}{x} \right)≤1[/MATH] so [MATH]x^2\sin^2 \left(\frac{1}{x} \right)≤x^2[/MATH] and the question becomes trivial.
Click to expand...
Oh yeah yeah, using properties of limits
 
lex said:
Presumably, you mean, 'prove this'.
[MATH]\sin^2 \left(\frac{1}{x} \right)≤1[/MATH] so [MATH]x^2\sin^2 \left(\frac{1}{x} \right)≤x^2[/MATH] and the question becomes trivial.
Click to expand...
Yes.
[MATH]\sin^2 \left(\frac{1}{x} \right)≤1 \implies 0≤x^2\sin^2 \left(\frac{1}{x} \right)≤x^2[/MATH]
and [MATH]\lim_{x \rightarrow 0} x^2 = 0 [/MATH][MATH]\therefore \lim_{x \rightarrow 0} x^2\sin^2 \left(\frac{1}{x} \right)[/MATH] exists and equals 0
 
lex said:
Yes.
[MATH]\sin^2 \left(\frac{1}{x} \right)≤1 \implies 0≤x^2\sin^2 \left(\frac{1}{x} \right)≤x^2[/MATH]
and [MATH]\lim_{x \rightarrow 0} x^2 = 0 [/MATH][MATH]\therefore \lim_{x \rightarrow 0} x^2\sin^2 \left(\frac{1}{x} \right)[/MATH] exists and equals 0
Click to expand...
Thank u for helping!!¡¡(*˘︶˘*).｡*♡
 
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