Evaluate lim x→∞ (3x+1)^(4/x), using l'hospital's rule as needed. show all work using proper notation. as you show your work, if a limit results in indeterminate form, state the type of indeterminate form before proceeding.

Respuesta :

[tex]\displaystyle\lim_{x\to\inty}(3x+1)^{4/x}=\lim_{x\to\infty}e^{\ln(3x+1)^{4/x}}=e^{\lim\limits_{x\to\infty}\ln(3x+1)^{4/x}}[/tex]

[tex]\displaystyle\lim_{x\to\infty}\ln(3x+1)^{4/x}=\lim_{x\to\infty}\frac{4\ln(3x+1)}x\stackrel{\mathrm{LHR}}=\lim_{x\to\infty}\frac{4\frac3{3x+1}}1=\lim_{x\to\infty}\frac{12}{3x+1}=0[/tex]

[tex]\implies\displaystyle\lim_{x\to\infty}(3x+1)^{4/x}=e^0=1[/tex]