Question

Question #67981

Answer 1

`lim_(x -> 0) (e^(3x) - e^(x))/x = 2`
`lim_(x -> 0) (e^(3x) - ex)/x = oo`

Explanation:

If you didn't make a typo and you meant `lim_(x -> 0) (e^(3x) - ex)/x` Then there is nothing else to do but substitute: `(1-0)/0 -> oo`

If you meant `lim_(x -> 0) (e^(3x) - e^(x))/x` Then we have an indetermination `(1-1) / 0 = 0/0` And we can use Bernoulli l'Hospital:

`lim_(x -> 0) (e^(3x) - e^(x))/x = lim_(x -> 0) ((d/dx)(e^(3x) - e^(x)))/((d/dx)x) = lim_(x -> 0) (3e^(3x) - e^(x))/1 = 2`

Answer 2

`2.`

Explanation:

We use this Standard Form of Limit : `lim_(x to 0)(e^x-1)/x=1.`

This will help us evaluate the reqd. limit without L'Hospital's Rule.

Reqd. Limit`=lim_(x to 0)(e^(3x)-e^x)/x`

`=lim_(x to 0) {((e^x)(e^(2x)-1))/x}`

`={lim_(x to 0) e^x}[2{lim_((2x) to 0) (e^(2x)-1)/(2x)}]`

`=(e^0)[2{1}]`

`:." The Reqd. Lim.="2.`

Enjoy Maths.!