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The plan is to evaluate the following limit: $$\lim\limits_{x\rightarrow 0} \frac{\frac{x}{7} +\frac{11}{x^3}}{\frac{x^2}{2} -\frac{2}{x^3}}.$$

So firstly there's no need really to use l'hopitals rule since we can multiply top and bottom by $x^3$ and then 'sub in' for zero and the limit is well defined, and gives $-11/2$.

If however we want to apply l'hopitals rule can we? So if you keep applying it without question as to whether it's valid, you arrive at the same answer.

However examining the numerator and denominator, neither limit exists separately, which is usually required for l'hopitals rule, and for it to give an indeterminate form of 0/0, $\infty/\infty$ etc.

I was talking to someone who mentioned possibly considering the left and right hand limits separately, so $x\rightarrow 0^+$ and $x\rightarrow 0^-$, in which case the numerator and denominator do have well defined limits which give indeterminate forms. Is this a way to proceed?

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  • $\begingroup$ Yes, this sounds like the best way to proceed. $\endgroup$ – Sinister Cutlass Nov 6 '15 at 0:25
  • $\begingroup$ In general, the best way to proceed is to not use L'Hospital. Taylor at order $1$ is equivalent, and it doesn't not have its drawbacks. $\endgroup$ – Bernard Nov 6 '15 at 0:34
  • $\begingroup$ this is a $\infty/\infty$ case, so in very general L'Hospital is applicable, although disadvantageously as you stated. $\endgroup$ – Michael Medvinsky Nov 6 '15 at 1:01
  • $\begingroup$ You can apply L'Hospital's rule repeatedly, but it will never provide an answer by itself. After three applications, both numerator and denominator will reduce to a single fraction that can be easily simplified to $-11/2$. But without that simplification, similar to the one you are foregoing at the start, applying L'Hospital's rule just takes you from one indeterminant form to another. The rule has never been guaranteed to be useful. $\endgroup$ – Paul Sinclair Nov 6 '15 at 2:59

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