Question:
Evaluate the following limits:
$\lim _{x \rightarrow 0} \frac{(1-\cos 4 x)}{(1-\cos 6 x)}$
Solution:
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, $\lim _{x \rightarrow 0} \frac{1-\cos 4 x}{1-\cos 6 x}=\frac{4}{9}$