Math  /  Calculus

Questionb. limx0(sinxx)1x3\lim _{x \rightarrow 0}\left(\frac{\sin x}{x}\right)^{\frac{1}{x^{3}}}

Studdy Solution

STEP 1

1. We are dealing with a limit problem involving an indeterminate form.
2. We will use properties of limits and logarithms to simplify the expression.
3. The known limit limx0sinxx=1\lim_{x \to 0} \frac{\sin x}{x} = 1 will be useful.

STEP 2

1. Recognize the indeterminate form and apply logarithms.
2. Simplify using the known limit.
3. Exponentiate to find the original limit.

STEP 3

Recognize that the expression (sinxx)1x3\left(\frac{\sin x}{x}\right)^{\frac{1}{x^{3}}} is of the form 11^\infty as x0x \to 0. To handle this, take the natural logarithm of the expression:
Let L=limx0(sinxx)1x3 L = \lim_{x \to 0} \left(\frac{\sin x}{x}\right)^{\frac{1}{x^3}} .
Then, take the natural logarithm:
lnL=limx01x3ln(sinxx) \ln L = \lim_{x \to 0} \frac{1}{x^3} \ln \left(\frac{\sin x}{x}\right)

STEP 4

Use the known limit limx0sinxx=1\lim_{x \to 0} \frac{\sin x}{x} = 1. Therefore, as x0x \to 0, ln(sinxx)ln1=0\ln \left(\frac{\sin x}{x}\right) \to \ln 1 = 0.
Apply L'Hôpital's Rule to evaluate the limit since it is of the form 00\frac{0}{0}:
lnL=limx0ln(sinxx)x3 \ln L = \lim_{x \to 0} \frac{\ln \left(\frac{\sin x}{x}\right)}{x^3}
Differentiate the numerator and the denominator:
Numerator: ddx(ln(sinxx))=cosxxsinxx2\frac{d}{dx} \left(\ln \left(\frac{\sin x}{x}\right)\right) = \frac{\cos x \cdot x - \sin x}{x^2}
Denominator: ddx(x3)=3x2\frac{d}{dx}(x^3) = 3x^2
Apply L'Hôpital's Rule:
lnL=limx0cosxxsinxx23x2 \ln L = \lim_{x \to 0} \frac{\frac{\cos x \cdot x - \sin x}{x^2}}{3x^2}
Simplify:
lnL=limx0cosxxsinx3x4 \ln L = \lim_{x \to 0} \frac{\cos x \cdot x - \sin x}{3x^4}

STEP 5

Evaluate the limit by further simplification or applying L'Hôpital's Rule again if necessary. However, note that as x0x \to 0, cosx1\cos x \to 1 and sinxx\sin x \to x, making the numerator approach zero faster than the denominator:
lnL=limx0xx3x4=limx003x4=0 \ln L = \lim_{x \to 0} \frac{x - x}{3x^4} = \lim_{x \to 0} \frac{0}{3x^4} = 0

STEP 6

Exponentiate to find the original limit:
Since lnL=0\ln L = 0, we have:
L=e0=1 L = e^0 = 1
The value of the limit is:
1 \boxed{1}

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