Question

using l'hopitale's rule solve lim x->0 (e^x+3x)^1/x

Answer 1

You know the basics of L'Hôpital's rule?

Answer 2

yes

Answer 3

okay, so basically, if you have a limit
\[\lim_{x \rightarrow a}\frac{f(x)}{g(x)}\]
and\[f(x)\rightarrow0 \ and \ g(x)\rightarrow0\]OR
\[f(x)\rightarrow \infty \ and \ g(x) \rightarrow \infty\]
Then
\[\lim_{x \rightarrow a}\frac{f(x)}{g(x)}=\lim_{x \rightarrow a}\frac{f \ '(x)}{g \ '(x)}\]

Answer 4

Can you still follow?

Answer 5

Sorry about that, my bad...
Your question is of the form
\[\huge \lim_{x \rightarrow a}f(x)^{g(x)}\]
Where
\[f(x) \rightarrow 1 \ and \ g(x) \rightarrow \infty\]

Catch me so far?

Answer 6

yes

Answer 7

Were you taught how to do this directly?

Answer 8

yeah it was a while ago though

Answer 9

Well, what were you taught? Or you want me to go over it step by step?

Answer 10

to plug in the 0 for the values of x and if it was 0/0 then to find the derivative and plug in 0 again

Answer 11

Well, that's not straightforward here, as you have a function raised to an exponent which happens to be another function, and not a function divided by another function...

Let's go over how to do this, shall we?
First, we assign a value to the limit, say
\[\huge \lim_{x \rightarrow a}f(x)^{g(x)} = L\]

Answer 12

Then let's apply the natural logarithm on both sides:
\[\huge \ln \left(\lim_{x \rightarrow a}f(x)^{g(x)}\right)=\ln \ L\]

Answer 13

The natural logarithm is a continuous function, so it can go inside the limit, like so:
\[\huge \lim_{x \rightarrow a} \ \ln \left(f(x)^{g(x)}\right)=\ln \ L\]

Answer 14

ok

Answer 15

Now, one property of the natural logarithm, a very useful one at that, is :
\[\large \ln (a^b) = b \ln (a)\]
Now to make good use of it...

Answer 16

\[\huge \lim_{x \rightarrow a} \ g(x)\ln \left(f(x)\right)=\ln \ L\]
Catch me so far?

Answer 17

yeah

Answer 18

Well, however cumbersome, this can be rewritten as:
\[\huge \lim_{x \rightarrow a}\frac{\ln (f(x))}{\frac{1}{g(x)}}=\ln \ L\]

Answer 19

I need to know that you understand this step.

Answer 20

i do

Answer 21

Well, then, anyway.., we know that
\[\large f(x) \rightarrow 1 \ and \ g(x) \rightarrow \infty\]

In that case, I hope it's clear to you that
\[\large \ln(f(x)) \rightarrow 0 \ and \frac1{g(x)} \rightarrow 0\]

... right?

Answer 22

ok

Answer 23

Well, now you can use L'Hôpital's Rule on
\[\huge \lim_{x \rightarrow a}\frac{\ln (f(x))}{\frac{1}{g(x)}}\]
Since both the numerator and denominator go to zero. So you can evaluate its limit.
However, that's not quite the end of it yet...

Answer 24

Suppose ^that particular limit goes to some K.
Then \[\huge \lim_{x \rightarrow a}\frac{\ln (f(x))}{\frac{1}{g(x)}} = \ln \ L = K\]

Answer 25

Then, clearly
\[\large L = e^K\]

Answer 26

And it was L we wanted to find in the first place.

Answer 27

So our final form is :
\[\huge \lim_{x \rightarrow a}f(x)^{g(x)}=e^{\lim_{x \rightarrow a}\frac{\ln (f(x))}{\frac{1}{g(x)}}}\]

Answer 28

That's not very neat, so let's clean it a bit:
\[\huge \lim_{x \rightarrow a}f(x)^{g(x)} = e^K\]where
\[\huge K= \lim_{x \rightarrow a}\frac{\ln (f(x))}{\frac{1}{g(x)}}\]

Answer 29

And there you go; L'Hôpital's Rule for the case of the indeterminate form
\[\huge 1^\infty\]

Answer 30

Now, for your question, simply plug in
\[f(x)=e^x + 3x\]\[g(x)=\frac{1}{x}\]

Answer 31

okay thanks!