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Here, we show you a step-by-step solved example of limits of exponential functions. This solution was automatically generated by our smart calculator:
xβ0limβ(1+3sinx)x1β
2
Rewrite the limit using the identity: ax=exln(a)
xβ0limβ(ex1βln(1+3sin(x)))
ο Passi intermedi
Multiplying the fraction by ln(1+3sin(x))
xβ0limβ(ex1ln(1+3sin(x))β)
Any expression multiplied by 1 is equal to itself
xβ0limβ(exln(1+3sin(x))β)
3
Multiplying the fraction by ln(1+3sin(x))
xβ0limβ(exln(1+3sin(x))β)
4
Apply the power rule of limits: xβalimβf(x)g(x)=xβalimβf(x)xβalimβg(x)
(xβ0limβ(e))limxβ0β(xln(1+3sin(x))β)
5
The limit of a constant is just the constant
elimxβ0β(xln(1+3sin(x))β)
ο Passi intermedi
Plug in the value 0 into the limit
xβ0limβ(0ln(1+3sin(0))β)
The sine of 0 equals 0
xβ0limβ(0ln(1+3β 0)β)
Multiply 3 times 0
xβ0limβ(0ln(1+0)β)
Add the values 1 and 0
xβ0limβ(0ln(1)β)
Calculating the natural logarithm of 1
xβ0limβ(00β)
6
If we directly evaluate the limit limxβ0β(xln(1+3sin(x))β) as x tends to 0, we can see that it gives us an indeterminate form
00β
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
xβ0limβ(dxdβ(x)dxdβ(ln(1+3sin(x)))β)
ο Passi intermedi
Find the derivative of the numerator
dxdβ(ln(1+3sin(x)))
The derivative of the natural logarithm of a function is equal to the derivative of the function divided by that function. If f(x)=lna (where a is a function of x), then fβ²(x)=aaβ²β
1+3sin(x)1βdxdβ(1+3sin(x))
The derivative of a sum of two or more functions is the sum of the derivatives of each function
1+3sin(x)1βdxdβ(3sin(x))
The derivative of a function multiplied by a constant is equal to the constant times the derivative of the function
3(1+3sin(x)1β)dxdβ(sin(x))
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 fβ²(x)=cos(x)β Dxβ(x)
3(1+3sin(x)1β)cos(x)
Multiplying the fraction by 3cos(x)
1+3sin(x)3cos(x)β
Find the derivative of the denominator
dxdβ(x)
The derivative of the linear function is equal to 1
1
Any expression divided by one (1) is equal to that same expression
elimxβ0β(1+3sin(x)3cos(x)β)
8
After deriving both the numerator and denominator, and simplifying, the limit results in
elimxβ0β(1+3sin(x)3cos(x)β)
ο Passi intermedi
Evaluate the limit xβ0limβ(1+3sin(x)3cos(x)β) by replacing all occurrences of x by 0
e1+3sin(0)3cos(0)β
The sine of 0 equals 0
e1+3β 03cos(0)β
Multiply 3 times 0
e1+03cos(0)β
Add the values 1 and 0
e13cos(0)β
The cosine of 0 equals 1
e13β 1β
Multiply 3 times 1
e13β
Divide 3 by 1
e3
9
Evaluate the limit limxβ0β(1+3sin(x)3cos(x)β) by replacing all occurrences of x by 0
e3
ξ Risposta finale al problema
e3ξ
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