Question

Evaluate: $\underset{x\to 0}{lim}\frac{cos2x-1}{cosx-1}$

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Answer 1

We have $\underset{x\to 0}{lim}\frac{cos2x-1}{cosx-1}$

At x=0, the value of the given function takes the form $\frac{0}{0}$

Now,

$=\underset{x\to 0}{lim}\frac{cos2x-1}{cosx-1}$

$=\underset{x\to 0}{lim}\frac{1-2si{n}^{2}x-1}{1-2si{n}^2\frac{x}{2}-1}$ $\{cos2x=1-2si{n}^{2}x\}$

$=\underset{x\to 0}{lim}\frac{si{n}^{2}x}{si{n}^2\left(\frac{x}{2}\right)}$

$=\underset{x\to 0}{lim}\frac{\left(\frac{sin{x}^2}{x}\right)\times x^2}{{\left(\frac{sin\left(\frac{x}{2}\right)}{\left(\frac{x}{2}\right)}\right)}^2\times \frac{x^2}{4}}$

$=\frac{4\underset{x\to 0}{lim}\frac{(sinx{)}^2}{x}}{\underset{x\to 0}{lim}{\left(\frac{sin\left(\frac{x}{2}\right)}{\left(\frac{x}{2}\right)}\right)}^2}$ $\{\underset{x\to 0}{lim}\frac{sinx}{x}=1\}$

$=\frac{4\times 1}{1}=4$