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

`\lim_(x\rightarrow \infty)((5x-3)/(5x+2))^(5x+1)=e^(-5)`

Explanation:

We are given that

`\lim_(x\rightarrow \infty)((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)=
`(5x-3)/(5x+2)`

Substitute the values in the formula

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

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

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

`y=e^(-5)`

`(1)/(\infty)=0`