If we directly evaluate the limit $\lim_{x\to0}\left(\frac{1-\cos\left(x\right)}{x^2}\right)$ as $x$ tends to $0$, we can see that it gives us an indeterminate form
We can solve this limit by applying L'Hôpital's rule, which consists of calculating the derivative of both the numerator and the denominator separately
After deriving both the numerator and denominator, and simplifying, the limit results in
If we directly evaluate the limit $\lim_{x\to0}\left(\frac{\sin\left(x\right)}{2x}\right)$ as $x$ tends to $0$, we can see that it gives us an indeterminate form
We can solve this limit by applying L'Hôpital's rule, which consists of calculating the derivative of both the numerator and the denominator separately
After deriving both the numerator and denominator, and simplifying, the limit results in
Evaluate the limit $\lim_{x\to0}\left(\frac{\cos\left(x\right)}{2}\right)$ by replacing all occurrences of $x$ by $0$
The cosine of $0$ equals $1$
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