Evaluate the following limits:


Evaluate the following limits:

$\lim _{x \rightarrow 0} \frac{(1-\cos 4 x)}{(1-\cos 6 x)}$



To Find: Limits

NOTE: First Check the form of imit. Used this method if the limit is satisfying any one from 7 indeterminate form.

In this Case, indeterminate Form is $\frac{0}{0}$

Formula used: $\lim _{x \rightarrow 0} \frac{1-\cos x}{x^{2}}=\frac{1}{2}$

Divide numerator and denominator by $x^{2}$, we have

So, by using the above formula, we have

$\lim _{x \rightarrow 0} \frac{1-\cos 4 x}{1-\cos 6 x}=\lim _{x \rightarrow 0} \frac{\frac{16[1-\cos 4 x]}{(4 x)^{2}}}{\frac{36[1-\cos 6 x]}{(6 x)^{2}}}=\frac{\frac{16}{2}}{\frac{36}{2}}=\frac{8}{18}=\frac{4}{9}$

Therefore, $\lim _{x \rightarrow 0} \frac{1-\cos 4 x}{1-\cos 6 x}=\frac{4}{9}$


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