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I was actually asked the following question that defied my understanding of the L´Hospital rule. If we have $\lim_{x\to 0} \frac{\sin(x)}{x}$ we compute the derivative which equals the $\cos(x)$ and then replace x by 0 which gives one. So $\lim_{x\to 0} \frac{\sin(x)}{x}=1$. The explanation would be that the ratio $\frac{\sin(x)}{x}$ would be the same as $\cos(x)$. But then I was asked why does the L´hopital rule does not work for $\lim_{x\to 5} \frac{\sin(x)}{x}$. I have not come to a consistent answer. Please help me out. Thanks!

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Its simple because at $x=5$ its not $\frac{0}{0}$ or $ \frac{\infty}{\infty} $ ( an indeterminate form which we cannot evaluate)

If the function is of $\frac{0}{0}$ or $ \frac{\infty}{\infty} $ type, then L Hopital Rule can be used. Else not. Otherwise also the limit the at $x \to 5$ exists. If you need a proof you will first need to be well acquainted with Mean Value Theorems of Lagrange and Cauchy ( Taylor Series would help).

Intuitively what L Hopital Rule does is that it approximates the function with its derivative .

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  • $\begingroup$ Yes I would need a proof. Thanks. $\endgroup$ – Pedro Gomes Mar 4 '17 at 19:10
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    $\begingroup$ @PedroGomes Have it here. Go to the proof section ( you will easily understand the general case ) en.m.wikipedia.org/wiki/L'H%C3%B4pital's_rule $\endgroup$ – Shashaank Mar 4 '17 at 19:13
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    $\begingroup$ @PedroGomes or here math.stackexchange.com/a/505788/333392 $\endgroup$ – Shashaank Mar 4 '17 at 19:17
  • $\begingroup$ @Shashaank Just FYI ... we don't require the numerator approach infinity; merely $\frac{\text{anythin}}{\infty}$ is suitable provided that the other conditions for LHR are met. $\endgroup$ – Mark Viola Mar 5 '17 at 0:17
  • $\begingroup$ @Dr.MV But every where we apply LHR we see that either it is 0/0 or infinity/infinity or a form reducible to it. If denominator is tending to 0 while numerator is not then limit doesn't exist. Please correct me if I am wrong because I have read this only . $\endgroup$ – Shashaank Mar 5 '17 at 4:58
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For $$\lim_{x\to5} \frac{\sin(x)}{x}$$

taking L'Hopital does not work since we are not at an indeterminate form.

Thus, $$\lim_{x\to5} \frac{\sin(x)}{x} = \frac{\sin(5)}{5}$$

Notice that in the case of $$\lim_{x\to0} \frac{\sin(x)}{x}$$

we have:

$$\frac{0}{0}$$

which is interminate, and so we apply the rule. In the limit approaches $5$ example, it is not.

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When limit is $x\to5$ then it is not indeterminate form so we cannot use L' Hospital rule there as L' Hospital rule is used for indeterminate form of limits not for normal limits.
I hope you've understood.

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