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Calcolatrice di Limiti di funzioni esponenziali

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limxβ†’0(1+3sinx)1x 
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1

Here, we show you a step-by-step solved example of limits of exponential functions. This solution was automatically generated by our smart calculator:

lim⁑xβ†’0(1+3sinx)1x\lim_{x\to0}\left(1+3sinx\right)^{\frac{1}{x}}
2

Rewrite the limit using the identity: ax=exln⁑(a)a^x=e^{x\ln\left(a\right)}

lim⁑xβ†’0(e1xln⁑(1+3sin⁑(x)))\lim_{x\to0}\left(e^{\frac{1}{x}\ln\left(1+3\sin\left(x\right)\right)}\right)

Multiplying the fraction by ln⁑(1+3sin⁑(x))\ln\left(1+3\sin\left(x\right)\right)

lim⁑xβ†’0(e1ln⁑(1+3sin⁑(x))x)\lim_{x\to0}\left(e^{\frac{1\ln\left(1+3\sin\left(x\right)\right)}{x}}\right)

Any expression multiplied by 11 is equal to itself

lim⁑xβ†’0(eln⁑(1+3sin⁑(x))x)\lim_{x\to0}\left(e^{\frac{\ln\left(1+3\sin\left(x\right)\right)}{x}}\right)
3

Multiplying the fraction by ln⁑(1+3sin⁑(x))\ln\left(1+3\sin\left(x\right)\right)

lim⁑xβ†’0(eln⁑(1+3sin⁑(x))x)\lim_{x\to0}\left(e^{\frac{\ln\left(1+3\sin\left(x\right)\right)}{x}}\right)
4

Apply the power rule of limits: lim⁑xβ†’af(x)g(x)=lim⁑xβ†’af(x)lim⁑xβ†’ag(x)\displaystyle{\lim_{x\to a}f(x)^{g(x)} = \lim_{x\to a}f(x)^{\displaystyle\lim_{x\to a}g(x)}}

(lim⁑xβ†’0(e))lim⁑xβ†’0(ln⁑(1+3sin⁑(x))x){\left(\lim_{x\to0}\left(e\right)\right)}^{\lim_{x\to0}\left(\frac{\ln\left(1+3\sin\left(x\right)\right)}{x}\right)}
5

The limit of a constant is just the constant

elim⁑xβ†’0(ln⁑(1+3sin⁑(x))x)e^{\lim_{x\to0}\left(\frac{\ln\left(1+3\sin\left(x\right)\right)}{x}\right)}

Plug in the value 00 into the limit

lim⁑xβ†’0(ln⁑(1+3sin⁑(0))0)\lim_{x\to0}\left(\frac{\ln\left(1+3\sin\left(0\right)\right)}{0}\right)

The sine of 00 equals 00

lim⁑xβ†’0(ln⁑(1+3β‹…0)0)\lim_{x\to0}\left(\frac{\ln\left(1+3\cdot 0\right)}{0}\right)

Multiply 33 times 00

lim⁑xβ†’0(ln⁑(1+0)0)\lim_{x\to0}\left(\frac{\ln\left(1+0\right)}{0}\right)

Add the values 11 and 00

lim⁑xβ†’0(ln⁑(1)0)\lim_{x\to0}\left(\frac{\ln\left(1\right)}{0}\right)

Calculating the natural logarithm of 11

lim⁑xβ†’0(00)\lim_{x\to0}\left(\frac{0}{0}\right)
6

If we directly evaluate the limit lim⁑xβ†’0(ln⁑(1+3sin⁑(x))x)\lim_{x\to0}\left(\frac{\ln\left(1+3\sin\left(x\right)\right)}{x}\right) as xx tends to 00, we can see that it gives us an indeterminate form

00\frac{0}{0}
7

We can solve this limit by applying L'HΓ΄pital's rule, which consists of calculating the derivative of both the numerator and the denominator separately

lim⁑xβ†’0(ddx(ln⁑(1+3sin⁑(x)))ddx(x))\lim_{x\to 0}\left(\frac{\frac{d}{dx}\left(\ln\left(1+3\sin\left(x\right)\right)\right)}{\frac{d}{dx}\left(x\right)}\right)

Find the derivative of the numerator

ddx(ln⁑(1+3sin⁑(x)))\frac{d}{dx}\left(\ln\left(1+3\sin\left(x\right)\right)\right)

The derivative of the natural logarithm of a function is equal to the derivative of the function divided by that function. If f(x)=lnβ€…af(x)=ln\:a (where aa is a function of xx), then fβ€²(x)=aβ€²a\displaystyle f'(x)=\frac{a'}{a}

11+3sin⁑(x)ddx(1+3sin⁑(x))\frac{1}{1+3\sin\left(x\right)}\frac{d}{dx}\left(1+3\sin\left(x\right)\right)

The derivative of a sum of two or more functions is the sum of the derivatives of each function

11+3sin⁑(x)ddx(3sin⁑(x))\frac{1}{1+3\sin\left(x\right)}\frac{d}{dx}\left(3\sin\left(x\right)\right)

The derivative of a function multiplied by a constant is equal to the constant times the derivative of the function

3(11+3sin⁑(x))ddx(sin⁑(x))3\left(\frac{1}{1+3\sin\left(x\right)}\right)\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){f(x) = \sin(x)}, then fβ€²(x)=cos⁑(x)β‹…Dx(x){f'(x) = \cos(x)\cdot D_x(x)}

3(11+3sin⁑(x))cos⁑(x)3\left(\frac{1}{1+3\sin\left(x\right)}\right)\cos\left(x\right)

Multiplying the fraction by 3cos⁑(x)3\cos\left(x\right)

3cos⁑(x)1+3sin⁑(x)\frac{3\cos\left(x\right)}{1+3\sin\left(x\right)}

Find the derivative of the denominator

ddx(x)\frac{d}{dx}\left(x\right)

The derivative of the linear function is equal to 11

11

Any expression divided by one (11) is equal to that same expression

elim⁑xβ†’0(3cos⁑(x)1+3sin⁑(x))e^{\lim_{x\to0}\left(\frac{3\cos\left(x\right)}{1+3\sin\left(x\right)}\right)}
8

After deriving both the numerator and denominator, and simplifying, the limit results in

elim⁑xβ†’0(3cos⁑(x)1+3sin⁑(x))e^{\lim_{x\to0}\left(\frac{3\cos\left(x\right)}{1+3\sin\left(x\right)}\right)}

Evaluate the limit lim⁑xβ†’0(3cos⁑(x)1+3sin⁑(x))\lim_{x\to0}\left(\frac{3\cos\left(x\right)}{1+3\sin\left(x\right)}\right) by replacing all occurrences of xx by 00

e3cos⁑(0)1+3sin⁑(0)e^{\frac{3\cos\left(0\right)}{1+3\sin\left(0\right)}}

The sine of 00 equals 00

e3cos⁑(0)1+3β‹…0e^{\frac{3\cos\left(0\right)}{1+3\cdot 0}}

Multiply 33 times 00

e3cos⁑(0)1+0e^{\frac{3\cos\left(0\right)}{1+0}}

Add the values 11 and 00

e3cos⁑(0)1e^{\frac{3\cos\left(0\right)}{1}}

The cosine of 00 equals 11

e3β‹…11e^{\frac{3\cdot 1}{1}}

Multiply 33 times 11

e31e^{\frac{3}{1}}

Divide 33 by 11

e3e^{3}
9

Evaluate the limit lim⁑xβ†’0(3cos⁑(x)1+3sin⁑(x))\lim_{x\to0}\left(\frac{3\cos\left(x\right)}{1+3\sin\left(x\right)}\right) by replacing all occurrences of xx by 00

e3e^{3}

ξ ƒ Risposta finale al problema

e3e^{3} ξ ƒ

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