Mister Exam
Other calculators:
-
How to use it?
- Limit of the function:
-
Limit of (-9+x^2)/(-3+x^2-2*x)
-
Limit of (-1+cos(x))/x^2
-
Limit of sin(17*x)/sin(12*x)
-
Limit of (5+x)/(-5+x^2+4*x)
-
Identical expressions
- (- one +cos(x))/x^ two
- ( minus 1 plus co sinus of e of (x)) divide by x squared
- ( minus one plus co sinus of e of (x)) divide by x to the power of two
- (-1+cos(x))/x2
- -1+cosx/x2
- (-1+cos(x))/x²
- (-1+cos(x))/x to the power of 2
- -1+cosx/x^2
- (-1+cos(x)) divide by x^2
-
Similar expressions
- (1+cos(x))/x^2
- (-1-cos(x))/x^2
- (-1+cosx)/x^2
Limit of the function (-1+cos(x))/x^2
The solution
You have entered
[src]
/-1 + cos(x)\
lim |-----------|
x->0+| 2 |
\ x /
$$\lim_{x \to 0^+}\left(\frac{\cos{\left(x \right)} - 1}{x^{2}}\right)$$
Limit((-1 + cos(x))/x^2, x, 0)
Detail solution
Let's take the limit
$$\lim_{x \to 0^+}\left(\frac{\cos{\left(x \right)} - 1}{x^{2}}\right)$$
We use the trigonometric function
transform
$$\lim_{x \to 0^+}\left(\frac{\cos{\left(x \right)} - 1}{x^{2}}\right) = \lim_{x \to 0^+}\left(- \frac{2 \sin^{2}{\left(\frac{x}{2} \right)}}{x^{2}}\right)$$
=
$$- 2 \left(\lim_{x \to 0^+}\left(\frac{\sin{\left(\frac{x}{2} \right)}}{x}\right)\right)^{2}$$
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(\frac{x}{2} \right)}}{x}\right) = \lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{2 u}\right)$$
=
$$\frac{\lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right)}{2}$$
The limit
$$\lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right)$$
is first remarkable limit, is equal to 1.
then
$$- 2 \left(\lim_{x \to 0^+}\left(\frac{\sin{\left(\frac{x}{2} \right)}}{x}\right)\right)^{2} = - 2 \left(\frac{\lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right)}{2}\right)^{2}$$
=
$$- \frac{2}{4}$$
=
$$- \frac{1}{2}$$
The final answer:
$$\lim_{x \to 0^+}\left(\frac{\cos{\left(x \right)} - 1}{x^{2}}\right) = - \frac{1}{2}$$
$$\lim_{x \to 0^+}\left(\frac{\cos{\left(x \right)} - 1}{x^{2}}\right)$$
We use the trigonometric function
sin(a)^2 = (1 - cos(2*a))/2
transform
$$\lim_{x \to 0^+}\left(\frac{\cos{\left(x \right)} - 1}{x^{2}}\right) = \lim_{x \to 0^+}\left(- \frac{2 \sin^{2}{\left(\frac{x}{2} \right)}}{x^{2}}\right)$$
=
$$- 2 \left(\lim_{x \to 0^+}\left(\frac{\sin{\left(\frac{x}{2} \right)}}{x}\right)\right)^{2}$$
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(\frac{x}{2} \right)}}{x}\right) = \lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{2 u}\right)$$
=
$$\frac{\lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right)}{2}$$
The limit
$$\lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right)$$
is first remarkable limit, is equal to 1.
then
$$- 2 \left(\lim_{x \to 0^+}\left(\frac{\sin{\left(\frac{x}{2} \right)}}{x}\right)\right)^{2} = - 2 \left(\frac{\lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right)}{2}\right)^{2}$$
=
$$- \frac{2}{4}$$
=
$$- \frac{1}{2}$$
The final answer:
$$\lim_{x \to 0^+}\left(\frac{\cos{\left(x \right)} - 1}{x^{2}}\right) = - \frac{1}{2}$$
Lopital's rule
We have indeterminateness of type
i.e. limit for the numerator is
$$\lim_{x \to 0^+}\left(\cos{\left(x \right)} - 1\right) = 0$$
and limit for the denominator is
$$\lim_{x \to 0^+} x^{2} = 0$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to 0^+}\left(\frac{\cos{\left(x \right)} - 1}{x^{2}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \left(\cos{\left(x \right)} - 1\right)}{\frac{d}{d x} x^{2}}\right)$$
=
$$\lim_{x \to 0^+}\left(- \frac{\sin{\left(x \right)}}{2 x}\right)$$
=
$$\lim_{x \to 0^+}\left(- \frac{\sin{\left(x \right)}}{2 x}\right)$$
=
$$- \frac{1}{2}$$
It can be seen that we have applied Lopital's rule (we have taken derivatives with respect to the numerator and denominator) 1 time(s)
0/0,
i.e. limit for the numerator is
$$\lim_{x \to 0^+}\left(\cos{\left(x \right)} - 1\right) = 0$$
and limit for the denominator is
$$\lim_{x \to 0^+} x^{2} = 0$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to 0^+}\left(\frac{\cos{\left(x \right)} - 1}{x^{2}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \left(\cos{\left(x \right)} - 1\right)}{\frac{d}{d x} x^{2}}\right)$$
=
$$\lim_{x \to 0^+}\left(- \frac{\sin{\left(x \right)}}{2 x}\right)$$
=
$$\lim_{x \to 0^+}\left(- \frac{\sin{\left(x \right)}}{2 x}\right)$$
=
$$- \frac{1}{2}$$
It can be seen that we have applied Lopital's rule (we have taken derivatives with respect to the numerator and denominator) 1 time(s)
The graph
One‐sided limits
[src]
/-1 + cos(x)\
lim |-----------|
x->0+| 2 |
\ x /
$$\lim_{x \to 0^+}\left(\frac{\cos{\left(x \right)} - 1}{x^{2}}\right)$$
-1/2
$$- \frac{1}{2}$$
= -0.5
/-1 + cos(x)\
lim |-----------|
x->0-| 2 |
\ x /
$$\lim_{x \to 0^-}\left(\frac{\cos{\left(x \right)} - 1}{x^{2}}\right)$$
-1/2
$$- \frac{1}{2}$$
= -0.5
= -0.5
Other limits x→0, -oo, +oo, 1
$$\lim_{x \to 0^-}\left(\frac{\cos{\left(x \right)} - 1}{x^{2}}\right) = - \frac{1}{2}$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{\cos{\left(x \right)} - 1}{x^{2}}\right) = - \frac{1}{2}$$
$$\lim_{x \to \infty}\left(\frac{\cos{\left(x \right)} - 1}{x^{2}}\right) = 0$$
More at x→oo
$$\lim_{x \to 1^-}\left(\frac{\cos{\left(x \right)} - 1}{x^{2}}\right) = -1 + \cos{\left(1 \right)}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{\cos{\left(x \right)} - 1}{x^{2}}\right) = -1 + \cos{\left(1 \right)}$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{\cos{\left(x \right)} - 1}{x^{2}}\right) = 0$$
More at x→-oo
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{\cos{\left(x \right)} - 1}{x^{2}}\right) = - \frac{1}{2}$$
$$\lim_{x \to \infty}\left(\frac{\cos{\left(x \right)} - 1}{x^{2}}\right) = 0$$
More at x→oo
$$\lim_{x \to 1^-}\left(\frac{\cos{\left(x \right)} - 1}{x^{2}}\right) = -1 + \cos{\left(1 \right)}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{\cos{\left(x \right)} - 1}{x^{2}}\right) = -1 + \cos{\left(1 \right)}$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{\cos{\left(x \right)} - 1}{x^{2}}\right) = 0$$
More at x→-oo
The graph