Question #67981

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Answer 1 · 2017-04-21T05:18:55

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

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 · 2017-04-21T07:35:51

$2.$

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.!