Question

Use l'Hopital's rule to find the limit y->0 (sqrt(5y+1) -1)/y

Answer 1

Well, thefirst thing you do is make sure you actually get an indeterminant form by plugging in your limit. If you donot come get0/0, infinity/infinity, etc, then you cannot yet use l'hopitals rule. In this case you do get the indeterminant 0/0, so we can just apply the rule. So we need to separately take the derivative of the numerator and then divide it by the deriviative of the denominator. As you can see, the deriviative of the denominator will just be 1, which quickly eliminates any issue with getting an undefinied answer. Now it's just up to you to properly take the derivative of the numerator and then plug in your limit.

Answer 2

Assuming you are in a situation where you can use L'Hopital's Rule, like @Psymon said, then you can use this relationship (given by the rule, taken from Wikipedia). Pysmon put it into words, and here is the math-speak.\[\lim_{x\to c}\frac{f(x)}{g(x)} = \lim_{x\to c}\frac{f'(x)}{g'(x)}\] http://en.wikipedia.org/wiki/L%27hospital%27s_rule

Answer 3

It's just making sure that when you do what is shown above that you take the derivative of the numerator SEPARATELY from the derivative of the denominator. A lot of people make the mistake of just doing a quotient rule when that is incorrect. The separate derivaitve of the numerator divided by the separate derivative of the denominator.

Answer 4

Good point! Just find some \(f(x)\div g(x)\) as your function that you are finding the limit of. Then, under the proper conditions, L'Hopital's Rule says that the \(\sf limit\) of \(f(x)\div g(x)\) is equal to the \(\sf limit\) of \(f'(x)\div g'(x)\).

Just to show another notation in case you like it better... Under the proper conditions, L'Hopital's Rule says that the \(\sf limit\) of \(f(x)\div g(x)\) is equal to the \(\sf limit\) of \(\dfrac{df(x)}{dx}\div \dfrac{dg(x)}{dx}=\dfrac{df}{dx}\div\dfrac{dg}{dx}\).

Answer 5

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