Calculus Examples

Evaluate Using L'Hospital's Rule limit as x approaches negative infinity of x/(2x-3)
limx-x2x-3
Step 1
Evaluate the limit of the numerator and the limit of the denominator.
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Step 1.1
Take the limit of the numerator and the limit of the denominator.
limx-xlimx-2x-3
Step 1.2
The limit at negative infinity of a polynomial of odd degree whose leading coefficient is positive is negative infinity.
-limx-2x-3
Step 1.3
The limit at negative infinity of a polynomial of odd degree whose leading coefficient is positive is negative infinity.
--
Step 1.4
Infinity divided by infinity is undefined.
Undefined
--
Step 2
Since -- is of indeterminate form, apply L'Hospital's Rule. L'Hospital's Rule states that the limit of a quotient of functions is equal to the limit of the quotient of their derivatives.
limx-x2x-3=limx-ddx[x]ddx[2x-3]
Step 3
Find the derivative of the numerator and denominator.
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Step 3.1
Differentiate the numerator and denominator.
limx-ddx[x]ddx[2x-3]
Step 3.2
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=1.
limx-1ddx[2x-3]
Step 3.3
By the Sum Rule, the derivative of 2x-3 with respect to x is ddx[2x]+ddx[-3].
limx-1ddx[2x]+ddx[-3]
Step 3.4
Evaluate ddx[2x].
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Step 3.4.1
Since 2 is constant with respect to x, the derivative of 2x with respect to x is 2ddx[x].
limx-12ddx[x]+ddx[-3]
Step 3.4.2
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=1.
limx-121+ddx[-3]
Step 3.4.3
Multiply 2 by 1.
limx-12+ddx[-3]
limx-12+ddx[-3]
Step 3.5
Since -3 is constant with respect to x, the derivative of -3 with respect to x is 0.
limx-12+0
Step 3.6
Add 2 and 0.
limx-12
limx-12
Step 4
Evaluate the limit of 12 which is constant as x approaches -.
12
 [x2  12  π  xdx ]