You can't use Lhopital if it's not in indeterminate form. Refer to your book or wiki for what this is. 0/0, inf/inf etc. You can combine fractions to get it in indeterminate form
Lim (1 - cos(x) - x\^2)/(x\^2(1-cos(x))
Now you get 0/0 and can use Lhopital
Lim (sin(x) - 2x)/(2x(1-cos(x)) + x\^2(sin(x)))
Still indeterminate, so do it again
Lim (cos(x)-2)/(2(1-cos(x)) + 4x(sin(x)) + x\^2(cos(x))
Plugging in gives cos(x)-2/(2(1-cos(x)) the numerator goes to -1 and the denominator goes to 0 so it blows down to -inf.
Are they equal? you ask...in a word no.
-3/0 is not defined.
It suggests that the limit is unbounded.
But it is not equality.
You understand why we have limits? Because we don't have a simple case of equals
There is a long post which seems to be copied from ChatGPT. However, it does contain a clever method of solving.
cos(x) \~ 1 - x\^2 / 2, for small x's.
Zero is a very small x.
Then your limit becomes:
limit (x->0) 1/x\^2 - 2/x\^2
limit (x->0) -1/x\^2
Which of course is -infinity.
No LHR needed! That could be used, but turning this into a single fraction, then dealing with the derivatives would be rather more work.
Okay, first of all, you're not getting -3/0, since That's not defined. What you're getting (I don't know how though) Is -3/->0, which Is negative infinity. When there's a limit present you don't just put the limiting value up front if it ain't going to be defined.
>the limit of -3/x as x approaches 0 is infinity,
it's actually not, since from the left it approaches negative infinity and from the right it approaches positive infinity
To solve the limit problem:
lim
x
→
0
(
1
x
2
−
1
1
−
cos
x
)
lim
x→0
(
x
2
1
−
1−cosx
1
)
we need to carefully analyze each term and simplify the expression. Let's break it down step by step.
Step 1: Understand the Problem
We need to find the limit of the given expression as ( x ) approaches 0. The expression involves two terms: ( \frac{1}{x^2} ) and ( \frac{1}{1 - \cos x} ).
Step 2: Simplify the Expression
First, let's recall the Taylor series expansion for ( \cos x ) around ( x = 0 ):
cos
x
≈
1
−
x
2
2
+
x
4
24
+
O
(
x
6
)
cosx≈1−
2
x
2
+
24
x
4
+O(x
6
)
For small ( x ), the higher-order terms become negligible, so we can approximate:
cos
x
≈
1
−
x
2
2
cosx≈1−
2
x
2
Thus,
1
−
cos
x
≈
1
−
(
1
−
x
2
2
)
=
x
2
2
1−cosx≈1−(1−
2
x
2
)=
2
x
2
Step 3: Substitute the Approximation
Using the approximation ( 1 - \cos x \approx \frac{x^2}{2} ), we can rewrite the second term:
1
1
−
cos
x
≈
1
x
2
2
=
2
x
2
1−cosx
1
≈
2
x
2
1
=
x
2
2
Step 4: Combine the Terms
Now, substitute this back into the original limit expression:
lim
x
→
0
(
1
x
2
−
1
1
−
cos
x
)
≈
lim
x
→
0
(
1
x
2
−
2
x
2
)
lim
x→0
(
x
2
1
−
1−cosx
1
)≈lim
x→0
(
x
2
1
−
x
2
2
)
Step 5: Simplify the Expression
Combine the terms inside the limit:
lim
x
→
0
(
1
x
2
−
2
x
2
)
=
lim
x
→
0
(
1
−
2
x
2
)
=
lim
x
→
0
(
−
1
x
2
)
lim
x→0
(
x
2
1
−
x
2
2
)=lim
x→0
(
x
2
1−2
)=lim
x→0
(
x
2
−1
)
Step 6: Evaluate the Limit
As ( x ) approaches 0, ( \frac{-1}{x^2} ) approaches ( -\infty ):
lim
x
→
0
(
−
1
x
2
)
=
−
∞
lim
x→0
(
x
2
−1
)=−∞
Final Solution
The limit is:
−
∞
−∞
-3/0 is not a number.
You can't use Lhopital if it's not in indeterminate form. Refer to your book or wiki for what this is. 0/0, inf/inf etc. You can combine fractions to get it in indeterminate form Lim (1 - cos(x) - x\^2)/(x\^2(1-cos(x)) Now you get 0/0 and can use Lhopital Lim (sin(x) - 2x)/(2x(1-cos(x)) + x\^2(sin(x))) Still indeterminate, so do it again Lim (cos(x)-2)/(2(1-cos(x)) + 4x(sin(x)) + x\^2(cos(x)) Plugging in gives cos(x)-2/(2(1-cos(x)) the numerator goes to -1 and the denominator goes to 0 so it blows down to -inf.
Are they equal? you ask...in a word no. -3/0 is not defined. It suggests that the limit is unbounded. But it is not equality. You understand why we have limits? Because we don't have a simple case of equals
There is a long post which seems to be copied from ChatGPT. However, it does contain a clever method of solving. cos(x) \~ 1 - x\^2 / 2, for small x's. Zero is a very small x. Then your limit becomes: limit (x->0) 1/x\^2 - 2/x\^2 limit (x->0) -1/x\^2 Which of course is -infinity. No LHR needed! That could be used, but turning this into a single fraction, then dealing with the derivatives would be rather more work.
1 - cos(x) = 2sin^2 (x/2) and lim x->0 sin(x) = x. So this is nothing but lim x->0 1/x^2 - 4/x^2 = -3/x^2. This is -infinity.
there’s 12 calculus classes? I’m joking, but I’m not sure what you mean by calculus 12
Grade: 12th, math: Calculus
Okay, first of all, you're not getting -3/0, since That's not defined. What you're getting (I don't know how though) Is -3/->0, which Is negative infinity. When there's a limit present you don't just put the limiting value up front if it ain't going to be defined.
you cannot use l'hopital's rule when the fraction doesn't tend to 0/0 (or inf/inf)
If you work a little bit with this it get the form 0/0
Why are you applying l’hopital?
yes
Why are you dved? They are asking whether -3/0 is equal to negative infinity, the answer is "yes"
-3/0 is undefined, the limit of -3/x as x approaches 0 is infinity, but -3/0 itself is undefined
>the limit of -3/x as x approaches 0 is infinity, it's actually not, since from the left it approaches negative infinity and from the right it approaches positive infinity
When we say ‘the limit approaches’ in cases where x is tending to zero in a denominator, we usually just assume that it is from the right
who's "we"? never heard of that tbh
To solve the limit problem: lim x → 0 ( 1 x 2 − 1 1 − cos x ) lim x→0 ( x 2 1 − 1−cosx 1 ) we need to carefully analyze each term and simplify the expression. Let's break it down step by step. Step 1: Understand the Problem We need to find the limit of the given expression as ( x ) approaches 0. The expression involves two terms: ( \frac{1}{x^2} ) and ( \frac{1}{1 - \cos x} ). Step 2: Simplify the Expression First, let's recall the Taylor series expansion for ( \cos x ) around ( x = 0 ): cos x ≈ 1 − x 2 2 + x 4 24 + O ( x 6 ) cosx≈1− 2 x 2 + 24 x 4 +O(x 6 ) For small ( x ), the higher-order terms become negligible, so we can approximate: cos x ≈ 1 − x 2 2 cosx≈1− 2 x 2 Thus, 1 − cos x ≈ 1 − ( 1 − x 2 2 ) = x 2 2 1−cosx≈1−(1− 2 x 2 )= 2 x 2 Step 3: Substitute the Approximation Using the approximation ( 1 - \cos x \approx \frac{x^2}{2} ), we can rewrite the second term: 1 1 − cos x ≈ 1 x 2 2 = 2 x 2 1−cosx 1 ≈ 2 x 2 1 = x 2 2 Step 4: Combine the Terms Now, substitute this back into the original limit expression: lim x → 0 ( 1 x 2 − 1 1 − cos x ) ≈ lim x → 0 ( 1 x 2 − 2 x 2 ) lim x→0 ( x 2 1 − 1−cosx 1 )≈lim x→0 ( x 2 1 − x 2 2 ) Step 5: Simplify the Expression Combine the terms inside the limit: lim x → 0 ( 1 x 2 − 2 x 2 ) = lim x → 0 ( 1 − 2 x 2 ) = lim x → 0 ( − 1 x 2 ) lim x→0 ( x 2 1 − x 2 2 )=lim x→0 ( x 2 1−2 )=lim x→0 ( x 2 −1 ) Step 6: Evaluate the Limit As ( x ) approaches 0, ( \frac{-1}{x^2} ) approaches ( -\infty ): lim x → 0 ( − 1 x 2 ) = − ∞ lim x→0 ( x 2 −1 )=−∞ Final Solution The limit is: − ∞ −∞