How do I solve these limits using L'Hospitals Rule

lim (1−(3/x))^x
x→∞

lim
x→1 ((3^x)−3)/((x^2)−1)

lim
x→∞ (arctan(x)/(1/x)−3)


I understand that these problems are indeterminate because they are 0/0 or inf/inf, but don't understand where I go from there to solve the problem. After I take the derivative, how would I go about doing these?

Thanks for the help!

lim (1−(3/x))^x
x→∞

For this one, you introduce ln(natural logarithm) to get that that exponent.

lim ln(1−(3/x))^x
x→∞

You put down the exponent.

lim xln(1−(3/x))
x→∞

Now you put the x as 1/x

lim ln(1−(3/x))
x→∞ 1/x

Now you have 1/∞ equal to zero, and you have 3/∞ equal to 0 and
Natural logarythm of 1 is equal to 0

We have now 0/0

And we can use lhopital rule.

Derivating

lim 3/x²/(1-(3/x))
x→∞ -1/x²

Rearranging.


lim 3/x²*(1-(3/x))
x→∞ -1/x²

Final steps.

lim -3x²
x→∞ x²*(1-(3/x))

Cancel x²

lim -3
x→∞ (1-(3/x))

So the limit goes to -3, but remember, we put natural logarithm to the limit, se we have to put euler to cancel it.

It would be something like this.

lim f(x) =ln(L)
x→K
And we have to know the value of L
We have to cancel the ln so. we put euler.

And the answer is e^-3 or 1/e³

Third limit does not need to use lhopital, the limit he always going to get indeterminate.

lim
x→∞ (arctan(x)/(1/x)−3)

We can make a little trick here, taking infinity values, arctan(x) grow slower than 1/x because it oscilates between [-1,1], 1/x grows slower, but faster than arctan(x), so this limit is going to be always infinity, no matter how manys you do lhopital rule.


and the second one.


lim
x→1 ((3^x)−3)/((x^2)−1)


So is indeterminate form 0/0

So we derivate.

The derivative of a constant with a function exponent is given by the rule.

k^x=k^x*lnk


so putting the limit like this, we use lhopital

lim ((3^x)−3)
x→1 x²-1


lim 3^x*ln3
x→1 2x

So we evaluate the limit now and it goes to.

.........3ln3
..........2

I hope it helps.

Apologize my bad english, im a mainly spanish talker.

Remember give the points to the better answer.