L' Hospital is very easy, is it?

Calculus Level 3

Please correct the 3rd statement as: Limit will oscillate b/w 0 and 2.

Note: I have said nothing about 3rd statement. It may be wrong or it may be right.

2 4 5 1

1 solution

Prakash Chandra Rai
Dec 10, 2014

L'Hospital rule states that if lim(x->a) f(x)/g(x) attains indeterminate form, then lim(x->a) f(x)/g(x) = lim(x->a) f'(x)/g'(x) if and only if R.H.S. exists.

Here existence means that it can yield finite as well as infinite value.

Recursion of L'Hospital is only valid when lim(x->a) f'(x)/g'(x) is also an indeterminate form.

Now coming to question. After applying L'Hospital Rule, we get (1-cos(x)) , when x->(infinity), we don't know the behaviour of cos(x) as x tends to infinity. So, we can't conclude anything about the RHS i.e(1-cos(x)) , so we can't evaluate this limit using L'Hospital Rule.

Hence, L'Hospital Rule is not applicable here.

You should rewrite the limit in statement 4 as $L = \lim_{x \rightarrow \infty} 1-\frac{\sin x}{x}$ since statement 5 means can be taken as either $L = 1$ or $\lim_{x \rightarrow \infty} \frac{\sin x}{x} = 1$ .

Jake Lai - 6 years, 6 months ago

In response to Jake Lai:

I am not able to understand your problem correctly. Please elaborate it.

What I understood is that you have problem in evaluating the below mentioned limit.

lim x->(infinity) sin(x)/(x) becomes 0 not 1.

Prakash Chandra Rai - 6 years, 6 months ago

Couldn't it be 1, like if you put x as 1/t in which t goes to zero 0, so that when you put lim t->0 you get sin(1/t)/1/t = 1?

Haries Ajrun - 6 years, 5 months ago

@Haries Ajrun – Since sin(x) will oscillate between -1 and 1 and in denominator we are tending to infinity. So limit will be 0.

Regarding your doubt, lim x->(zero) sin(x)/(x) = 1 . It is correct but it is only true for cases when argument of sin and denominator both tend to zero. This result fails even if they tend to a non zero number. And here, they are tending to infinity,so this result is not valid here.

Reply if you want a proof why it is only true for cases where argument of sin and denominator both tend to zero.

Note: Here you can't apply L'Hospital rule because numerator will oscillate b/w -1 and 1 and denominator is tending to infinity. No (zero)/(zero) form, so no L'Hospital.

Prakash Chandra Rai - 6 years, 5 months ago

L' Hospital is very easy, is it?

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