Certainly! Let's calculate the limit step-by-step:
Consider the limit:
[tex]\[
\lim_{x \to 0} \frac{\tan(5x) - \sin(5x)}{x^3}
\][/tex]
1. Express the functions in terms of series expansions:
For small [tex]\( x \)[/tex], we can expand [tex]\( \tan(5x) \)[/tex] and [tex]\( \sin(5x) \)[/tex] using their Taylor series expansions around [tex]\( x = 0 \)[/tex].
[tex]\[
\tan(5x) = 5x + \frac{(5x)^3}{3} + O(x^5)
\][/tex]
[tex]\[
\sin(5x) = 5x - \frac{(5x)^3}{6} + O(x^5)
\][/tex]
2. Subtract the series expansions:
[tex]\[
\tan(5x) - \sin(5x) = \left( 5x + \frac{(5x)^3}{3} + O(x^5) \right) - \left( 5x - \frac{(5x)^3}{6} + O(x^5) \right)
\][/tex]
Combine the like terms:
[tex]\[
\tan(5x) - \sin(5x) = \left( 5x - 5x \right) + \left( \frac{(5x)^3}{3} + \frac{(5x)^3}{6} \right) + O(x^5)
\][/tex]
Simplifying inside the parentheses:
[tex]\[
\tan(5x) - \sin(5x) = \left( \frac{125x^3}{3} + \frac{125x^3}{6} \right) + O(x^5)
\][/tex]
3. Simplify the coefficients:
[tex]\[
\left( \frac{125x^3}{3} + \frac{125x^3}{6} \right) = 125x^3 \left( \frac{1}{3} + \frac{1}{6} \right)
\][/tex]
Combine the fractions:
[tex]\[
\frac{1}{3} + \frac{1}{6} = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}
\][/tex]
So:
[tex]\[
\tan(5x) - \sin(5x) = 125x^3 \cdot \frac{1}{2} + O(x^5) = \frac{125x^3}{2} + O(x^5)
\][/tex]
4. Substitute back into the limit:
Now we substitute this into our original limit.
[tex]\[
\lim_{x \to 0} \frac{\tan(5x) - \sin(5x)}{x^3} = \lim_{x \to 0} \frac{\frac{125x^3}{2} + O(x^5)}{x^3}
\][/tex]
Simplifying the fraction:
[tex]\[
\lim_{x \to 0} \frac{\frac{125x^3}{2} + O(x^5)}{x^3} = \lim_{x \to 0} \left( \frac{125}{2} + \frac{O(x^5)}{x^3} \right)
\][/tex]
5. Evaluate the limit:
As [tex]\( x \)[/tex] approaches 0, the term [tex]\(\frac{O(x^5)}{x^3}\)[/tex] will approach 0.
Therefore:
[tex]\[
\lim_{x \to 0} \left( \frac{125}{2} + \frac{O(x^5)}{x^3} \right) = \frac{125}{2}
\][/tex]
Thus, the limit is:
[tex]\[
\boxed{\frac{125}{2}}
\][/tex]
