Why can't I apply L'Hopital's rule to all indeterminate forms?

[tmt1]

I have a certain set of problems (i.e. https://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/limcondirectory/LimitConstant.html), where many problems are in an indeterminate form ($\frac{0}{0}$) but if we apply L'Hopital's rule it yields an incorrect answer. Instead, I have to simplify the expression and then evaluate the expression like normal.

For example, $$\lim_{{x}\to{3}} \frac{x^4 - 81}{2x^2 - 5x - 3}$$.

If I apply l'hopital's rule, I get $\frac{x^3}{x} = 9$ but this is the wrong answer.

 

[I like Serena]

I have a certain set of problems (i.e. https://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/limcondirectory/LimitConstant.html), where many problems are in an indeterminate form ($\frac{0}{0}$) but if we apply L'Hopital's rule it yields an incorrect answer. Instead, I have to simplify the expression and then evaluate the expression like normal.

For example, $$\lim_{{x}\to{3}} \frac{x^4 - 81}{2x^2 - 5x - 3}$$.

If I apply l'hopital's rule, I get $\frac{x^3}{x} = 9$ but this is the wrong answer.

Hi tmt! ;)

Applying L'Hôpital's rule I'm getting:
$$\lim_{{x}\to{3}} \frac{x^4 - 81}{2x^2 - 5x - 3}
= \lim_{{x}\to{3}} \frac{4x^3}{4x - 5}
= \frac{4\cdot 3^3}{4\cdot 3 - 5}
= \frac{108}{7}
$$

 

[HallsofIvy]

Applying L'Hopital's rule, that is, differentiating numerator and denominator separately, we have \(\displaystyle \frac{4x^3}{4x- 5}\), not "\(\displaystyle \frac{x^3}{x}\). It looks like you forgot the "-5" in the derivative of the denominator.

Added later: for a problem like this, you don't really need to apply "L'Hopital's rule". Since numerator and denominator are polynomials that are equal to 0 when x= 3, it follows immediately that both have a factor of x- 3. $$x^4- 81= (x^2- 9)(x^2+ 9)= (x- 3)(x+ 3)(x^3+ 9)$$. $$2x^2- 5x- 3= (2x+ 1)(x- 3)$$. (Knowing that x- 3 is a factor makes that a lot easier to factor.)

So, for x not equal to 3, $$\frac{x^2- 81}{x^2- 5x- 3}=$$$$\frac{(x- 3)(x+ 3)(x^2+ 9)}{(x- 3)(2x+ 1)}$$$$= \frac{(x+ 3)(x^2+ 9)}{2x+ 1}$$ and so $$\lim_{x\to 3}\frac{x^2- 81}{x^2- 5x- 3}= \lim_{x\to 3} \frac{(x+ 3)(x^2+ 9)}{2x+ 1}$$$$= \frac{(3+ 3)(9+ 9)}{ 6+ 1}= \frac{108}{7}$$.