Math  /  Calculus

Questionlimx1exex1\lim _{x \rightarrow 1} \frac{e^{x}-e}{x-1}

Studdy Solution

STEP 1

1. The problem involves finding the limit of a function as x x approaches a specific value.
2. The expression exex1\frac{e^{x}-e}{x-1} is in an indeterminate form 00 \frac{0}{0} when x=1 x = 1 .
3. L'Hôpital's Rule can be applied to resolve the indeterminate form.

STEP 2

1. Verify the indeterminate form.
2. Apply L'Hôpital's Rule.
3. Calculate the derivative of the numerator and the denominator.
4. Evaluate the limit using the derivatives.

STEP 3

Verify the indeterminate form by substituting x=1 x = 1 into the expression:
Numerator: exe=e1e=ee=0 e^{x} - e = e^{1} - e = e - e = 0
Denominator: x1=11=0 x - 1 = 1 - 1 = 0
Since both the numerator and denominator are zero, the expression is in the indeterminate form 00 \frac{0}{0} .

STEP 4

Apply L'Hôpital's Rule, which states that for limits of the form 00 \frac{0}{0} or \frac{\infty}{\infty} , the limit of the ratio of the derivatives can be taken:
limx1exex1=limx1ddx(exe)ddx(x1) \lim _{x \rightarrow 1} \frac{e^{x}-e}{x-1} = \lim _{x \rightarrow 1} \frac{\frac{d}{dx}(e^{x}-e)}{\frac{d}{dx}(x-1)}

STEP 5

Calculate the derivative of the numerator and the denominator:
1. Derivative of the numerator exe e^{x} - e is ddx(ex)=ex \frac{d}{dx}(e^{x}) = e^{x} .
2. Derivative of the denominator x1 x - 1 is ddx(x)=1 \frac{d}{dx}(x) = 1 .

STEP 6

Evaluate the limit using the derivatives obtained:
limx1ex1=limx1ex=e1=e \lim _{x \rightarrow 1} \frac{e^{x}}{1} = \lim _{x \rightarrow 1} e^{x} = e^{1} = e
The limit is e \boxed{e} .

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