Question

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.

Answer 1

`\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)}`


`\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)(12)/(3x+1)=0`


`\implies\displaystyle\lim_(x\to\infty)(3x+1)^(4/x)=e^0=1`