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Sagot :
To solve the limit [tex]\(\lim _{x \rightarrow \pi / 2} \frac{\cos x}{\left(\frac{\pi}{2}-x\right)}\)[/tex], we need to carefully analyze the behavior of both the numerator and denominator as [tex]\( x \)[/tex] approaches [tex]\( \frac{\pi}{2} \)[/tex].
1. Behavior of the Numerator:
The numerator is [tex]\(\cos x\)[/tex]. As [tex]\( x \)[/tex] approaches [tex]\( \frac{\pi}{2} \)[/tex], [tex]\(\cos x\)[/tex] approaches [tex]\( \cos\left(\frac{\pi}{2}\right) \)[/tex], which is [tex]\(0\)[/tex]. Therefore, the numerator approaches [tex]\(0\)[/tex].
2. Behavior of the Denominator:
The denominator is [tex]\(\left(\frac{\pi}{2} - x\right)\)[/tex]. As [tex]\( x \)[/tex] approaches [tex]\( \frac{\pi}{2} \)[/tex], the expression [tex]\(\left(\frac{\pi}{2} - x\right)\)[/tex] approaches [tex]\(0\)[/tex].
3. Form of the Limit:
With both the numerator and the denominator approaching [tex]\(0\)[/tex], this limit is an indeterminate form of type [tex]\(\frac{0}{0}\)[/tex]. We can apply L'Hôpital's Rule to resolve this indeterminate form.
4. Applying L'Hôpital's Rule:
L'Hôpital's Rule states that for limits of the form [tex]\(\frac{0}{0}\)[/tex] or [tex]\(\frac{\infty}{\infty}\)[/tex]:
[tex]\[ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \][/tex]
provided the limit on the right exists. Here, we need to differentiate the numerator and the denominator.
- The derivative of the numerator [tex]\(\cos x\)[/tex] is [tex]\(-\sin x\)[/tex].
- The derivative of the denominator [tex]\(\frac{\pi}{2} - x\)[/tex] is [tex]\(-1\)[/tex].
Applying L'Hôpital's Rule:
[tex]\[ \lim_{x \to \frac{\pi}{2}} \frac{\cos x}{\frac{\pi}{2} - x} = \lim_{x \to \frac{\pi}{2}} \frac{-\sin x}{-1} = \lim_{x \to \frac{\pi}{2}} \sin x \][/tex]
5. Evaluating the Simplified Limit:
Now we need to evaluate [tex]\(\lim_{x \to \frac{\pi}{2}} \sin x\)[/tex]. As [tex]\( x \)[/tex] approaches [tex]\( \frac{\pi}{2} \)[/tex], [tex]\(\sin x\)[/tex] approaches [tex]\(\sin\left(\frac{\pi}{2}\right)\)[/tex], which is [tex]\(1\)[/tex].
Therefore, the limit is:
[tex]\[ \lim _{x \rightarrow \pi / 2} \frac{\cos x}{\left(\frac{\pi}{2}-x\right)} = 1 \][/tex]
1. Behavior of the Numerator:
The numerator is [tex]\(\cos x\)[/tex]. As [tex]\( x \)[/tex] approaches [tex]\( \frac{\pi}{2} \)[/tex], [tex]\(\cos x\)[/tex] approaches [tex]\( \cos\left(\frac{\pi}{2}\right) \)[/tex], which is [tex]\(0\)[/tex]. Therefore, the numerator approaches [tex]\(0\)[/tex].
2. Behavior of the Denominator:
The denominator is [tex]\(\left(\frac{\pi}{2} - x\right)\)[/tex]. As [tex]\( x \)[/tex] approaches [tex]\( \frac{\pi}{2} \)[/tex], the expression [tex]\(\left(\frac{\pi}{2} - x\right)\)[/tex] approaches [tex]\(0\)[/tex].
3. Form of the Limit:
With both the numerator and the denominator approaching [tex]\(0\)[/tex], this limit is an indeterminate form of type [tex]\(\frac{0}{0}\)[/tex]. We can apply L'Hôpital's Rule to resolve this indeterminate form.
4. Applying L'Hôpital's Rule:
L'Hôpital's Rule states that for limits of the form [tex]\(\frac{0}{0}\)[/tex] or [tex]\(\frac{\infty}{\infty}\)[/tex]:
[tex]\[ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \][/tex]
provided the limit on the right exists. Here, we need to differentiate the numerator and the denominator.
- The derivative of the numerator [tex]\(\cos x\)[/tex] is [tex]\(-\sin x\)[/tex].
- The derivative of the denominator [tex]\(\frac{\pi}{2} - x\)[/tex] is [tex]\(-1\)[/tex].
Applying L'Hôpital's Rule:
[tex]\[ \lim_{x \to \frac{\pi}{2}} \frac{\cos x}{\frac{\pi}{2} - x} = \lim_{x \to \frac{\pi}{2}} \frac{-\sin x}{-1} = \lim_{x \to \frac{\pi}{2}} \sin x \][/tex]
5. Evaluating the Simplified Limit:
Now we need to evaluate [tex]\(\lim_{x \to \frac{\pi}{2}} \sin x\)[/tex]. As [tex]\( x \)[/tex] approaches [tex]\( \frac{\pi}{2} \)[/tex], [tex]\(\sin x\)[/tex] approaches [tex]\(\sin\left(\frac{\pi}{2}\right)\)[/tex], which is [tex]\(1\)[/tex].
Therefore, the limit is:
[tex]\[ \lim _{x \rightarrow \pi / 2} \frac{\cos x}{\left(\frac{\pi}{2}-x\right)} = 1 \][/tex]
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