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# Find the limit of $\frac{1-\cos\left(x\right)}{x^2}$ as $x$ approaches 0

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$\frac{1}{2}$$\,\,\left(\approx 0.5\right) Got another answer? Verify it here ## Step-by-step Solution Problem to solve: \lim_{x\to 0}\left(\frac{1-\cos\left(x\right)}{x^2}\right) Choose the solving method Plug in the value 0 into the limit \frac{1-\cos\left(0\right)}{0^2} The cosine of 0 equals 1 \frac{1-1\cdot 1}{0^2} Multiply -1 times 1 \frac{1-1}{0^2} Subtract the values 1 and -1 \frac{0}{0^2} Calculate the power 0^2 \frac{0}{0} 1 If we directly evaluate the limit \lim_{x\to 0}\left(\frac{1-\cos\left(x\right)}{x^2}\right) as x tends to 0, we can see that it gives us an indeterminate form \frac{0}{0} 2 We can solve this limit by applying L'Hôpital's rule, which consists of calculating the derivative of both the numerator and the denominator separately \lim_{x\to 0}\left(\frac{\frac{d}{dx}\left(1-\cos\left(x\right)\right)}{\frac{d}{dx}\left(x^2\right)}\right) Find the derivative of the numerator \frac{d}{dx}\left(1-\cos\left(x\right)\right) The derivative of a sum of two or more functions is the sum of the derivatives of each function \frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(-\cos\left(x\right)\right) The derivative of the constant function (1) is equal to zero \frac{d}{dx}\left(-\cos\left(x\right)\right) The derivative of a function multiplied by a constant (-1) is equal to the constant times the derivative of the function -\frac{d}{dx}\left(\cos\left(x\right)\right) The derivative of the cosine of a function is equal to minus the sine of the function times the derivative of the function, in other words, if f(x) = \cos(x), then f'(x) = -\sin(x)\cdot D_x(x) \sin\left(x\right) Find the derivative of the denominator \frac{d}{dx}\left(x^2\right) The power rule for differentiation states that if n is a real number and f(x) = x^n, then f'(x) = nx^{n-1} 2x 3 After deriving both the numerator and denominator, the limit results in \lim_{x\to0}\left(\frac{\sin\left(x\right)}{2x}\right) Plug in the value 0 into the limit \frac{\sin\left(0\right)}{2\cdot 0} The sine of 0 equals 0 \frac{0}{2\cdot 0} Multiply 2 times 0 \frac{0}{0} 4 If we directly evaluate the limit \lim_{x\to 0}\left(\frac{\sin\left(x\right)}{2x}\right) as x tends to 0, we can see that it gives us an indeterminate form \frac{0}{0} 5 We can solve this limit by applying L'Hôpital's rule, which consists of calculating the derivative of both the numerator and the denominator separately \lim_{x\to 0}\left(\frac{\frac{d}{dx}\left(\sin\left(x\right)\right)}{\frac{d}{dx}\left(2x\right)}\right) Find the derivative of the numerator \frac{d}{dx}\left(\sin\left(x\right)\right) The derivative of the sine of a function is equal to the cosine of that function times the derivative of that function, in other words, if {f(x) = \sin(x)}, then {f'(x) = \cos(x)\cdot D_x(x)} \cos\left(x\right) Find the derivative of the denominator \frac{d}{dx}\left(2x\right) The derivative of the linear function times a constant, is equal to the constant 2 6 After deriving both the numerator and denominator, the limit results in \lim_{x\to0}\left(\frac{\cos\left(x\right)}{2}\right) 7 Evaluate the limit \lim_{x\to0}\left(\frac{\cos\left(x\right)}{2}\right) by replacing all occurrences of x by 0 \frac{\cos\left(0\right)}{2} The cosine of 0 equals 1 \frac{1}{2} Divide 1 by 2 \frac{1}{2} 8 Simplifying, we get \frac{1}{2} ## Final Answer \frac{1}{2}$$\,\,\left(\approx 0.5\right)$
SnapXam A2

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e
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ln
log
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lim
d/dx
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>=
<=
sin
cos
tan
cot
sec
csc

asin
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atan
acot
asec
acsc

sinh
cosh
tanh
coth
sech
csch

asinh
acosh
atanh
acoth
asech
acsch

### Tips on how to improve your answer:

$\lim_{x\to 0}\left(\frac{1-\cos\left(x\right)}{x^2}\right)$

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~ 0.05 s