The given limit problem is \(\displaystyle \lim_{x \to 0} \left(\frac{1}{x} – \frac{1}{\sin(x)}\right) \). In this post, we will learn how to solve this limit question.
Solution:
\( \displaystyle \lim_{x \to 0} \left(\frac{1}{x} – \frac{1}{\sin(x)}\right) = \lim_{x \to 0} \frac{\sin(x)\, – x}{x \sin(x)} \\ \)As \( x \to 0 \), both the numerator and denominator approach \( 0 \). This is a \( \frac{0}{0} \) indeterminate form, so we can apply L’Hôpital’s Rule.
Apply L’Hôpital’s Rule (Differentiate the numerator and denominator separately):
\( \displaystyle \lim_{x \to 0} \frac{\sin(x)\, – x}{x \sin(x)} \\ \displaystyle = \lim_{x \to 0} \frac{\cos(x)\, – 1}{\sin(x) + x \cos(x)} \\ \)This is still a \( \frac{0}{0} \) indeterminate form, so we apply L’Hôpital’s Rule again.
Apply L’Hôpital’s Rule a second time (Differentiate the numerator and denominator separately):
Now, the limit becomes:
\( \displaystyle \lim_{x \to 0} \frac{-\sin(x)}{\cos(x) + \cos(x)\,- x \sin(x)} \\ \displaystyle = \lim_{x \to 0} \frac{-\sin(x)}{2 \cos(x)\,- x \sin(x)} \\ \displaystyle = \frac{-\sin(0)}{2 \cos(0)\, – 0 \cdot \sin(0)} \\ \displaystyle = \frac{0}{2 \cdot 1\, – 0} = \frac{0}{2} = 0 \\ \)Therefore,
\[ \fbox{$ \lim_{x \to 0} \left(\frac{1}{x} – \frac{1}{\sin(x)}\right) = 0 $} \]Please let me know in the comments if you find any error in this solution.