Question

Trouble with another limit..

Answer 1

It is \[\lim_{x \rightarrow 0} \left( 1 + 3x \right)^{2+\frac{ 1 }{ 4x ^{2}}}\]

Answer 2

Answer is \[e ^{\frac{ 3 }{ 4 }}\], just dont know how to get there.

Answer 3

Really having touble with the \[\lim_{x \rightarrow 0} \frac{ \ln \left( 1 + 3x \right) }{\frac{ 1 }{ 2 + \frac{ 1 }{ 4x ^{2} } } }\] bit.

Answer 4

More Specifically, \[\frac{ d }{ dx } {\frac{ 1 }{ 2 + \frac{ 1 }{ 4x ^{2} } } } \]

Answer 5

the \[4x^{2}\] should just be a 4x.

Answer 6

do you know the answer to this problem?\[\lim_{n\rightarrow \infty}\left(1+\frac{3}{n}\right)^n\] Because a substitution of:\[x=\frac{1}{n}\]and noticing that\[x\rightarrow 0 \iff n\rightarrow \infty\] can change this problem into the first equation (with some extra baggage that can be taken care of on the side).

Answer 7

Answer would just be e^3 right? Clever, will give original a shot this way.

Answer 8

Just saw another condition, cant use L'Hoptials. This sucks.

Answer 9

The limit goes to infinity for your expression.
if your expression is \[ \lim_{x\to 0} (1 + 3x^2)^{\frac 1 {4x^2}}\]
then your limit is \( e^{3/4}\). Just use the definition of \( e\) as hinted by Joemath

Answer 10

Woops!! two different values,
http://www.wolframalpha.com/input/?i=limit+x-%3E0+%281%2B3*x%29^%282+%2B+1%2F%284*x^2%29%29

limit x->0- goes to 0 because, |1-3x| <1 while x->0+ goes to infinity because you would encounter something like e^(3/4 * 1/x)

Answer 11

Thanks experimentX, I'd made a mistake typing the above. Actually looks like this. limit x->0 (1+3*x)^(2 + 1/(4*x))

Answer 12

Well here's how wolfram does it.

Answer 13

yes, one way to deal it with is to use the definition of e. remove the square first, it evaluates to 1, the second term (1+3/x)^(4/x) is (1+3/x)^(3/x* 4/3) which is e^(4/3)

Answer 14

Awesome, thank you :) - wolframs way seems a bit long winded.