Sure, let's solve the limit step-by-step to find [tex]\( L = \lim_{{x \to 0}} \left(e^{3x} - 5x\right)^{1/x} \)[/tex].
First, the expression inside the limit is [tex]\((e^{3x} - 5x)^{1/x}\)[/tex]. To handle this, we can use the properties of logarithms and exponentials for limits.
1. Taking the natural logarithm:
Let's first take the natural logarithm of [tex]\( L \)[/tex].
[tex]\[
\ln L = \ln \left( \lim_{{x \to 0}} \left(e^{3x} - 5x\right)^{1/x} \right)
\][/tex]
2. Logarithm of a limit:
Using the property of logarithms, [tex]\(\ln (a^b) = b \ln (a)\)[/tex], we have:
[tex]\[
\ln L = \lim_{{x \to 0}} \frac{\ln (e^{3x} - 5x)}{x}
\][/tex]
Here, the limit and logarithm have been interchanged.
3. Expression analysis:
Let's rewrite the argument of the limit in a more workable form:
[tex]\[
\ln (e^{3x} - 5x)
\][/tex]
When [tex]\( x \)[/tex] is very close to 0, [tex]\( e^{3x} \approx 1 + 3x + \frac{9x^2}{2} \)[/tex], and thus:
[tex]\[
e^{3x} - 5x \approx 1 + 3x + \frac{9x^2}{2} - 5x = 1 - 2x + \frac{9x^2}{2}
\][/tex]
4. Logarithm expansion:
Using the approximation [tex]\(\ln(1 + y) \approx y\)[/tex] when [tex]\( y \)[/tex] is close to 0, we have:
[tex]\[
\ln(e^{3x} - 5x) \approx \ln\left(1 - 2x + \frac{9x^2}{2}\right)
\approx -2x + \frac{9x^2}{2}
\][/tex]
5. Simplifying the limit:
Plug the approximate value of the logarithm back into our limit expression:
[tex]\[
\ln L \approx \lim_{{x \to 0}} \frac{-2x + \frac{9x^2}{2}}{x}
= \lim_{{x \to 0}} \left( -2 + \frac{9x}{2} \right)
\][/tex]
As [tex]\( x \)[/tex] approaches 0, the term [tex]\(\frac{9x}{2}\)[/tex] becomes negligible:
[tex]\[
\ln L = -2
\][/tex]
6. Exponentiating to remove the logarithm:
Finally, to find [tex]\( L \)[/tex], we exponentiate both sides:
[tex]\[
L = e^{-2}
\][/tex]
Therefore, the limit is:
[tex]\[
L = \lim_{{x \to 0}} \left(e^{3x} - 5x\right)^{1/x} = e^{-2}
\][/tex]
So, the value of the limit is [tex]\( e^{-2} \)[/tex].
