0

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!

gymbvghjkgkjkhgfkl
  • 8,090
Pedro Gomes
  • 3,891

3 Answers3

1

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.

K Split X
  • 6,565
1

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 .

Dando18
  • 5,368
Shashaank
  • 903
  • Yes I would need a proof. Thanks. – Pedro Gomes Mar 04 '17 at 19:10
  • 1
    @PedroGomes Have it here. Go to the proof section ( you will easily understand the general case ) https://en.m.wikipedia.org/wiki/L'H%C3%B4pital's_rule – Shashaank Mar 04 '17 at 19:13
  • 1
    @PedroGomes or here http://math.stackexchange.com/a/505788/333392 – Shashaank Mar 04 '17 at 19:17
  • @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. – Mark Viola Mar 05 '17 at 00:17
  • @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 . – Shashaank Mar 05 '17 at 04:58
  • @shashaank The basic forms are $0/0$ and $\text{anything}/\infty$. The prevailing misconception is that in the latter, the numerator must also tend to $\infty$. But in fact, the limit of the numerator need not even exist for LHR to apply. – Mark Viola Mar 05 '17 at 05:04
  • @Dr.MV Ok Thank you for clarification. Can this fact be made seen in the proof of L Hopital using Cauchy's theorem. I will try to see and find out in the proof itself. If I cannot find out ' why that's right ' I will ask. These small facts make the big difference. Thank you for pointing it out. – Shashaank Mar 05 '17 at 05:09
  • This proof https://math.stackexchange.com/questions/505535/proof-of-lhospitals-rule/505788#505788 is for $\frac{0}{0}$. I hope that someone help me to find a proof for $\frac{\infty}{\infty}$. Thanks a lot. – Student Oct 02 '21 at 16:52
  • @Student if $\frac{f(x_0)}{g(x_0)}=\frac{0}{0}$ then $\frac{f(x_0)}{g(x_0)}= \frac{constant_1}{\infty}\frac{\infty}{constant_2}=\frac{\infty}{\infty}$…..actually wherever I have written $\infty$ I should technically write something tending to $\infty$ since $\infty$ in not in the set of Reals. But that is correct because $f(x) \rightarrow 0$ and not that $f(x)=0$ – Shashaank Oct 09 '21 at 20:28
0

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.

sgrmshrsm7
  • 815