# Trigonometry limit problem

### Trigonometry limit problem

Hi,
I have a problem which needs to be solved without using L'Hosiptal rule.
Can anyone help me?
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### Re: Trigonometry limit problem

The first thing I would do is let $$u= \sqrt[3]{cos(x)}$$. Then $$cos(x)= u^3$$, $$cos^2(x)= u^6$$, and $$sin^2(x)= 1- cos^2(x)= 1- u^6$$. Also, as x goes to 0, cos(x) goes to 1 so u goes to 1.

The problem becomes $$\lim_{u\to 1}\frac{u^3- u}{1- u^6}$$.

Factor numerator and denominator and cancel where you can.

HallsofIvy

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