L'Hopital's Rule for finding limits?

[twelvety]

Why is it that L'Hopitals rule does not work for:

sqrt(3x+1)/sqrt(x+1) as x --> infinity?

Also how would I find this limit using other means?

 

[Deleted member 4993]

Why is it that L'Hopitals rule does not work for:

sqrt(3x+1)/sqrt(x+1) as x --> infinity?

Also how would I find this limit using other means?

\(\displaystyle \frac{\sqrt{3x+1}}{\sqrt{x+1}} \, = \, {\sqrt\frac{3x+1}{x+1}} \, = \, {\sqrt\frac{3+\frac{1}{x}}{1+\frac{1}{x}} \,\)

L'Hospital's rule is not guaranteed to yield the limit.

 

[galactus]

Another thing you can notice is by dividing, we get:

\(\displaystyle \frac{\sqrt{3x+1}}{\sqrt{x+1}}=\sqrt{\frac{3x+1}{x+1}}=\sqrt{3-\frac{2}{x+1}}\)

Now, the limit is rather obvious.

No need for L'Hopital anyway.

 

[BigGlenntheHeavy]

lim ?(3x+1)/?(x+1) = ?(3x)/?(x) = ?[(3x)/(x)] = ?3.
x-> ?