Join the growing community of curious minds on IDNLearn.com and get the answers you need. Find accurate and detailed answers to your questions from our experienced and dedicated community members.
Sagot :
To find the limit of the expression [tex]\(\lim_{x \rightarrow 0}\left(\frac{1}{x^2}-\frac{1}{\sin^2 x}\right)\)[/tex], let's proceed step-by-step.
### Step 1: Understand the Behavior of [tex]\( \sin(x) \)[/tex] Near Zero
We know that as [tex]\( x \)[/tex] approaches 0, [tex]\(\sin(x)\)[/tex] behaves very similarly to [tex]\( x \)[/tex]. More precisely, we can use the small-angle approximation:
[tex]\[ \sin(x) \approx x \][/tex]
when [tex]\( x \)[/tex] is near 0.
### Step 2: Rewrite the Expression Using the Approximation
However, to accurately find the limit, we need to consider this approximation more precisely. One useful identity for small angles is:
[tex]\[ \sin(x) = x - \frac{x^3}{6} + \mathcal{O}(x^5) \][/tex]
where [tex]\( \mathcal{O}(x^5) \)[/tex] represents higher-order terms that become negligible much faster than [tex]\( x^3 \)[/tex].
### Step 3: Substitute [tex]\( \sin(x) \)[/tex] into the Expression
Let’s use our more precise approximation ([tex]\(\sin(x) \approx x - \frac{x^3}{6}\)[/tex]):
[tex]\[ \sin^2(x) = \left(x - \frac{x^3}{6}\right)^2 \approx x^2 - \frac{x^4}{3} \][/tex]
So:
[tex]\[ \frac{1}{\sin^2(x)} \approx \frac{1}{x^2 - \frac{x^4}{3}} \][/tex]
### Step 4: Simplify the Denominator for [tex]\( \frac{1}{\sin^2(x)} \)[/tex]
For small [tex]\( x \)[/tex], [tex]\( x^4 \)[/tex] is significantly smaller than [tex]\( x^2 \)[/tex], so we can approximate:
[tex]\[ \frac{1}{x^2 - \frac{x^4}{3}} \approx \frac{1}{x^2 \left(1 - \frac{x^2}{3}\right)} \][/tex]
Recalling the binomial expansion for [tex]\( (1 - y)^{-1} \)[/tex] for small [tex]\( y \)[/tex]:
[tex]\[ (1 - \frac{x^2}{3})^{-1} \approx 1 + \frac{x^2}{3} \][/tex]
Thus:
[tex]\[ \frac{1}{x^2 \left(1 - \frac{x^2}{3}\right)} \approx \frac{1}{x^2} \left(1 + \frac{x^2}{3}\right) = \frac{1}{x^2} + \frac{1}{3} \][/tex]
### Step 5: Substitute and Combine Terms
Now substitute the approximation into the original limit expression:
[tex]\[ \frac{1}{x^2} - \frac{1}{\sin^2(x)} \approx \frac{1}{x^2} - \left(\frac{1}{x^2} + \frac{1}{3}\right) = \frac{1}{x^2} - \frac{1}{x^2} - \frac{1}{3} = -\frac{1}{3} \][/tex]
### Step 6: Take the Limit
Since all approximations hold as [tex]\( x \)[/tex] approaches 0, the remaining term is simply:
[tex]\[ \lim_{x \to 0} \left(\frac{1}{x^2} - \frac{1}{\sin^2(x)}\right) = -\frac{1}{3} \][/tex]
Therefore, the limit is:
[tex]\[ \boxed{-\frac{1}{3}} \][/tex]
### Step 1: Understand the Behavior of [tex]\( \sin(x) \)[/tex] Near Zero
We know that as [tex]\( x \)[/tex] approaches 0, [tex]\(\sin(x)\)[/tex] behaves very similarly to [tex]\( x \)[/tex]. More precisely, we can use the small-angle approximation:
[tex]\[ \sin(x) \approx x \][/tex]
when [tex]\( x \)[/tex] is near 0.
### Step 2: Rewrite the Expression Using the Approximation
However, to accurately find the limit, we need to consider this approximation more precisely. One useful identity for small angles is:
[tex]\[ \sin(x) = x - \frac{x^3}{6} + \mathcal{O}(x^5) \][/tex]
where [tex]\( \mathcal{O}(x^5) \)[/tex] represents higher-order terms that become negligible much faster than [tex]\( x^3 \)[/tex].
### Step 3: Substitute [tex]\( \sin(x) \)[/tex] into the Expression
Let’s use our more precise approximation ([tex]\(\sin(x) \approx x - \frac{x^3}{6}\)[/tex]):
[tex]\[ \sin^2(x) = \left(x - \frac{x^3}{6}\right)^2 \approx x^2 - \frac{x^4}{3} \][/tex]
So:
[tex]\[ \frac{1}{\sin^2(x)} \approx \frac{1}{x^2 - \frac{x^4}{3}} \][/tex]
### Step 4: Simplify the Denominator for [tex]\( \frac{1}{\sin^2(x)} \)[/tex]
For small [tex]\( x \)[/tex], [tex]\( x^4 \)[/tex] is significantly smaller than [tex]\( x^2 \)[/tex], so we can approximate:
[tex]\[ \frac{1}{x^2 - \frac{x^4}{3}} \approx \frac{1}{x^2 \left(1 - \frac{x^2}{3}\right)} \][/tex]
Recalling the binomial expansion for [tex]\( (1 - y)^{-1} \)[/tex] for small [tex]\( y \)[/tex]:
[tex]\[ (1 - \frac{x^2}{3})^{-1} \approx 1 + \frac{x^2}{3} \][/tex]
Thus:
[tex]\[ \frac{1}{x^2 \left(1 - \frac{x^2}{3}\right)} \approx \frac{1}{x^2} \left(1 + \frac{x^2}{3}\right) = \frac{1}{x^2} + \frac{1}{3} \][/tex]
### Step 5: Substitute and Combine Terms
Now substitute the approximation into the original limit expression:
[tex]\[ \frac{1}{x^2} - \frac{1}{\sin^2(x)} \approx \frac{1}{x^2} - \left(\frac{1}{x^2} + \frac{1}{3}\right) = \frac{1}{x^2} - \frac{1}{x^2} - \frac{1}{3} = -\frac{1}{3} \][/tex]
### Step 6: Take the Limit
Since all approximations hold as [tex]\( x \)[/tex] approaches 0, the remaining term is simply:
[tex]\[ \lim_{x \to 0} \left(\frac{1}{x^2} - \frac{1}{\sin^2(x)}\right) = -\frac{1}{3} \][/tex]
Therefore, the limit is:
[tex]\[ \boxed{-\frac{1}{3}} \][/tex]
We appreciate your contributions to this forum. Don't forget to check back for the latest answers. Keep asking, answering, and sharing useful information. For trustworthy answers, rely on IDNLearn.com. Thanks for visiting, and we look forward to assisting you again.
