Answer :
To solve the limit [tex]\(\lim _{x \rightarrow \alpha} \frac{x \cot \alpha-\alpha \cot x}{x-\alpha}\)[/tex], we can follow these steps:
1. Identify the form of the expression: We notice that both the numerator and the denominator approach [tex]\(0\)[/tex] as [tex]\(x \to \alpha\)[/tex]. This suggests that we might use L'Hôpital's Rule, which applies when we have an indeterminate form such as [tex]\(\frac{0}{0}\)[/tex].
2. Apply L'Hôpital's Rule: According to L'Hôpital's Rule, if [tex]\(\lim_{x \to a} \frac{f(x)}{g(x)}\)[/tex] results in the indeterminate form [tex]\(\frac{0}{0}\)[/tex] or [tex]\(\frac{\infty}{\infty}\)[/tex], then:
[tex]\[ \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \][/tex]
provided the limit on the right-hand side exists.
3. Differentiate the numerator and the denominator:
The numerator [tex]\(N(x)\)[/tex] is [tex]\(x \cot \alpha - \alpha \cot x\)[/tex], and:
[tex]\[ N'(x) = \frac{d}{dx} \left( x \cot \alpha - \alpha \cot x \right) \][/tex]
Given that [tex]\(\cot \alpha\)[/tex] is a constant with respect to [tex]\(x\)[/tex]:
[tex]\[ N'(x) = \cot \alpha - \alpha \frac{d}{dx} (\cot x) \][/tex]
Since:
[tex]\[ \frac{d}{dx} (\cot x) = - \csc^2 x \][/tex]
Thus:
[tex]\[ N'(x) = \cot \alpha + \alpha \csc^2 x \][/tex]
The denominator [tex]\(D(x)\)[/tex] is [tex]\(x - \alpha\)[/tex], and:
[tex]\[ D'(x) = \frac{d}{dx} (x - \alpha) = 1 \][/tex]
4. Use L'Hôpital's Rule and find the limit:
[tex]\[ \lim_{x \to \alpha} \frac{x \cot \alpha - \alpha \cot x}{x - \alpha} = \lim_{x \to \alpha} \frac{\cot \alpha + \alpha \csc^2 x}{1} \][/tex]
Next, we substitute [tex]\(x = \alpha\)[/tex] into the derivative:
[tex]\[ \lim_{x \to \alpha} (\cot \alpha + \alpha \csc^2 x) = \cot \alpha + \alpha \csc^2 \alpha \][/tex]
Since [tex]\(\csc \alpha = \frac{1}{\sin \alpha}\)[/tex], then [tex]\(\csc^2 \alpha = \frac{1}{\sin^2 \alpha}\)[/tex]:
[tex]\[ \lim_{x \to \alpha} (\cot \alpha + \alpha \csc^2 x) = \cot \alpha + \alpha \frac{1}{\sin^2 \alpha} \][/tex]
Furthermore, [tex]\(\cot \alpha = \frac{\cos \alpha}{\sin \alpha}\)[/tex], so:
[tex]\[ \cot \alpha + \alpha \frac{1}{\sin^2 \alpha} = \frac{\cos \alpha}{\sin \alpha} + \frac{\alpha}{\sin^2 \alpha} \][/tex]
Multiplying throughout by [tex]\(\sin^2 \alpha\)[/tex]:
[tex]\[ \frac{\cos \alpha \sin \alpha + \alpha}{\sin^2 \alpha} \][/tex]
5. Simplify the expression:
Combining the fractions yields:
[tex]\[ \frac{\cos \alpha \sin \alpha + \alpha}{\sin^2 \alpha} = \frac{\alpha \sin(\alpha) + \cos(\alpha) \sin(\alpha)}{\sin^2(\alpha)} \][/tex]
Now recognize that [tex]\(\tan \alpha = \frac{\sin \alpha}{\cos \alpha}\)[/tex], thus:
[tex]\[ = \frac{\alpha \sin(\alpha) + \cos (\alpha)^3 \tan(\alpha)^2}{\cos(\alpha)^3 \tan(\alpha)^3} \][/tex]
Therefore, the limit is given by:
[tex]\[ \boxed{\left(\frac{\alpha \sin(\alpha) + \cos(\alpha)^3 \tan(\alpha)^2}{\cos(\alpha)^3 \tan(\alpha)^3}\right)} \][/tex]
1. Identify the form of the expression: We notice that both the numerator and the denominator approach [tex]\(0\)[/tex] as [tex]\(x \to \alpha\)[/tex]. This suggests that we might use L'Hôpital's Rule, which applies when we have an indeterminate form such as [tex]\(\frac{0}{0}\)[/tex].
2. Apply L'Hôpital's Rule: According to L'Hôpital's Rule, if [tex]\(\lim_{x \to a} \frac{f(x)}{g(x)}\)[/tex] results in the indeterminate form [tex]\(\frac{0}{0}\)[/tex] or [tex]\(\frac{\infty}{\infty}\)[/tex], then:
[tex]\[ \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \][/tex]
provided the limit on the right-hand side exists.
3. Differentiate the numerator and the denominator:
The numerator [tex]\(N(x)\)[/tex] is [tex]\(x \cot \alpha - \alpha \cot x\)[/tex], and:
[tex]\[ N'(x) = \frac{d}{dx} \left( x \cot \alpha - \alpha \cot x \right) \][/tex]
Given that [tex]\(\cot \alpha\)[/tex] is a constant with respect to [tex]\(x\)[/tex]:
[tex]\[ N'(x) = \cot \alpha - \alpha \frac{d}{dx} (\cot x) \][/tex]
Since:
[tex]\[ \frac{d}{dx} (\cot x) = - \csc^2 x \][/tex]
Thus:
[tex]\[ N'(x) = \cot \alpha + \alpha \csc^2 x \][/tex]
The denominator [tex]\(D(x)\)[/tex] is [tex]\(x - \alpha\)[/tex], and:
[tex]\[ D'(x) = \frac{d}{dx} (x - \alpha) = 1 \][/tex]
4. Use L'Hôpital's Rule and find the limit:
[tex]\[ \lim_{x \to \alpha} \frac{x \cot \alpha - \alpha \cot x}{x - \alpha} = \lim_{x \to \alpha} \frac{\cot \alpha + \alpha \csc^2 x}{1} \][/tex]
Next, we substitute [tex]\(x = \alpha\)[/tex] into the derivative:
[tex]\[ \lim_{x \to \alpha} (\cot \alpha + \alpha \csc^2 x) = \cot \alpha + \alpha \csc^2 \alpha \][/tex]
Since [tex]\(\csc \alpha = \frac{1}{\sin \alpha}\)[/tex], then [tex]\(\csc^2 \alpha = \frac{1}{\sin^2 \alpha}\)[/tex]:
[tex]\[ \lim_{x \to \alpha} (\cot \alpha + \alpha \csc^2 x) = \cot \alpha + \alpha \frac{1}{\sin^2 \alpha} \][/tex]
Furthermore, [tex]\(\cot \alpha = \frac{\cos \alpha}{\sin \alpha}\)[/tex], so:
[tex]\[ \cot \alpha + \alpha \frac{1}{\sin^2 \alpha} = \frac{\cos \alpha}{\sin \alpha} + \frac{\alpha}{\sin^2 \alpha} \][/tex]
Multiplying throughout by [tex]\(\sin^2 \alpha\)[/tex]:
[tex]\[ \frac{\cos \alpha \sin \alpha + \alpha}{\sin^2 \alpha} \][/tex]
5. Simplify the expression:
Combining the fractions yields:
[tex]\[ \frac{\cos \alpha \sin \alpha + \alpha}{\sin^2 \alpha} = \frac{\alpha \sin(\alpha) + \cos(\alpha) \sin(\alpha)}{\sin^2(\alpha)} \][/tex]
Now recognize that [tex]\(\tan \alpha = \frac{\sin \alpha}{\cos \alpha}\)[/tex], thus:
[tex]\[ = \frac{\alpha \sin(\alpha) + \cos (\alpha)^3 \tan(\alpha)^2}{\cos(\alpha)^3 \tan(\alpha)^3} \][/tex]
Therefore, the limit is given by:
[tex]\[ \boxed{\left(\frac{\alpha \sin(\alpha) + \cos(\alpha)^3 \tan(\alpha)^2}{\cos(\alpha)^3 \tan(\alpha)^3}\right)} \][/tex]
