Evaluate Using L'Hospital's Rule limit as x approaches 0 of (e^x+x)^(1/x)

Problem

(lim_x→0)(e^x+x)

Solution

1. Identify the indeterminate form. As x→0 the base (e^x+x)→1 and the exponent 1/x→∞ resulting in the indeterminate form 1^∞

2. Rewrite the limit using the natural logarithm to transform the expression into a form suitable for L'Hospital's Rule. Let y=(e^x+x)^1/x then ln(y)=ln(e^x+x)/x

3. Evaluate the limit of the logarithm. We need to find L=(lim_x→0)(ln(e^x+x)/x) This is a 0/0 indeterminate form.

4. Apply L'Hospital's Rule by differentiating the numerator and the denominator with respect to x

L=(lim_x→0)(d(ln(e^x+x))/d(x)/d(x)/d(x))

5. Differentiate the expressions.

L=(lim_x→0)((e^x+1)/(e^x+x)/1)

6. Substitute x=0 into the simplified limit.

L=(e^0+1)/(e^0+0)

L=(1+1)/1

L=2

7. Exponentiate the result to find the original limit. Since (lim_x→0)(ln(y))=2 then (lim_x→0)(y)=e^2

Final Answer

(lim_x→0)(e^x+x)=e^2

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