Answer :
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]