Question

Find the limit. Use l'Hospital's Rule if appropriate. If there is a more elementary method, consider using it. lim x→[infinity] (5x − 3/5x + 2)^ 5x + 1

Answer 1

Answer:

$\lim_{x\rightarrow \infty}(\frac{5x-3}{5x+2})^{5x+1}=e^{-5}$

Step-by-step explanation:

We are given that

$\lim_{x\rightarrow \infty}(\frac{5x-3}{5x+2})^{5x+1}$

When substitute limit x tends to infinity then it is $1^{\infty}$ which is indeterminant form

Wheny=$\lim_{x\rightarrow \infty} f(x)^{g(x))$

and $1^{\infty}$

Then use $y=e^{\lim_{x\rightarrow \infty} g(x)(f(x)-1)}$

We have g(x)=5x+1 and f(x)=$\frac{5x-3}{5x+2}$

Substitute the values in the formula

$y=e^{\lim_{x\rightarrow \infty}(5x+1)(\frac{5x-3}{5x+2}-1)}$

$y=e^{\lim_{x\rightarrow \infty}(\frac{-5(5x+1)}{5x+2})}$

$y=e^{\lim_{x\rightarrow \infty}(\frac{-5(1+\frac{1}{5x})}{1+\frac{2}{5x}})}$

$y=e^{-5}$

$\frac{1}{\infty}=0$