Finding the limit as $x$ approaches zero for $\left(x+e^{2x}\right)^{\frac3x}$

Question

Finding the limit as $x$ approaches zero of

$$\left(x+e^{2x}\right)^{\frac3x}$$

I honestly don't know what to do here.

Don't know if we can use the l'Hopital rule but i doubt.

Answer 1

Just compose Taylor series $$y=\left(x+e^{2 x}\right)^{\frac 3x}\implies \log(y)=\frac 3x \log\left(x+e^{2 x}\right)$$ $$x+e^{2x}=1+3 x+2 x^2+O\left(x^3\right)$$ $$\log\left(x+e^{2 x}\right)=3 x-\frac{5 x^2}{2}+O\left(x^3\right)$$ $$\log(y)=9-\frac{15 x}{2}+O\left(x^2\right)$$ $$y=e^{\log(y)}=e^9\left(1-\frac{15 x}{2} \right)+O\left(x^2\right)$$ which shows the limit and how it is approached.

Answer 2

Hint: $(x+e^{2x})^{3/x}= \exp(\frac{3}{x} \ln (x+e^{2x})).$

Now use L'Hospital for $\frac{3}{x} \ln (x+e^{2x}$ with $x \to 0.$

Answer 3

HINT

$$\left(x+e^{2x}\right)^{\frac3x}=\left(e^{2x}\right)^{\frac3x}\left[\left(1+\frac x{e^{2x}}\right)^{\frac{e^{2x}}x}\right]^{\frac 3{e^{2x}}}$$

Answer 4

$$L=\lim_{x \to 0} [x+e^{2x}]^{3/x}\lim_{x \to 0} [1+3x+2x^2+..]^{3/x}=\exp[\lim_{x \to 0}\frac{3}{x}[1+3x+2x^2-1]]=e^9.$$

Edit for OP When $L=\lim_{x \to a} f(x)^{g(x)} \rightarrow 1^{\infty}.$ Then $$L=\exp[\lim_{x\to a} g(x)\ln (1+f(x)-1)=\exp[\lim_{x \to a} g(x)(f(x)-1)]$$ Also use $e^{2x}=1+2x+2x^2+...$

Answer 5

$$L=\lim_{x \to 0} (x+e^{2x})^{3/x}$$ $$\dfrac {\ln L}3=\lim_{x \to 0}\dfrac 1x \ln (x+e^{2x})$$ $$\dfrac {\ln L}3=\lim_{x \to 0}\dfrac 1x \ln e^{2x}(1+\dfrac x{e^{2x}})$$ $$\dfrac {\ln L}3=2+\lim_{x \to 0}\dfrac 1x \ln \left( 1+\dfrac x{e^{2x}}\right)$$ $$\dfrac {\ln L}3=2+\lim_{x \to 0}\dfrac 1 {e^{2x}}\dfrac {e^{2x}}x \ln\left( 1+\dfrac x{e^{2x}}\right)$$

Note that: $$\lim_{y \to 0}\dfrac {\ln (1+y)}y=1$$ $$\lim_{x \to 0}\dfrac 1 {e^{2x}}=1$$ So that: $$\dfrac {\ln L}3=2+1 \times 1 \implies L=e^9$$