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Use l'Hopital's rule to find the exact value of

[math]\lim_{x \to 0} \left(e^{\frac{5}{x}}-3x\right)^{\frac{x}{2}}[/math] .

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This is what I have done so far:

[math]\lim_{x \to 0}\ln y=\lim_{x \to 0}\ln\left(e^{\frac{5}{x}}-3x\right)^{\frac{x}{2}}[/math][math]\lim_{x \to 0}\ln y=\lim_{x \to 0}\frac{x\ln\left(e^{\frac{5}{x}}-3x\right)}{2}[/math][math]\lim_{x \to 0}\frac{x\ln\left(e^{\frac{5}{x}}-3x\right)}{2} = 0[/math][math]\lim_{x \to 0}\left(e^{\frac{5}{x}}-3x\right)^{\frac{x}{2}} = \lim_{x \to 0}y = \lim_{x \to 0} e^{\ln y} =e^{0} =1[/math]

The answer key to the worksheet says the answer is actually

**[math]e^{2.5}[/math], not 1**which is what I and the rest of my classmates got. Is the answer key wrong or am I missing something. Thanks!