How do you use L'hospital's rule to find the limit #lim_(x->oo)xsin(pi/x)# ?

Answer 1

To use L'Hôpital's Rule to find the limit lim(x→∞) x*sin(π/x), follow these steps:

  1. Identify the indeterminate form of the limit as x approaches infinity, which is 0*∞.

  2. Rewrite the limit as a fraction: lim(x→∞) sin(π/x) / (1/x).

  3. Take the derivatives of the numerator and denominator separately.

  4. Evaluate the derivatives and substitute them back into the original limit expression.

  5. Repeat steps 3 and 4 as needed until the limit is no longer in an indeterminate form.

  6. Evaluate the limit using the new expression.

Remember that L'Hôpital's Rule can only be applied when both the numerator and denominator approach either 0 or infinity as x approaches the limit.

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Answer 2
I assume that the limit is at #+infty# (otherwise the same arguments can be easily adapted to the #-infty# case). So #lim_{x to +infty} x sin(pi / x)#.
In this case you have an indeterminate form of type #+infty * 0#. In fact, the first factor is simply #x to +infty# and the second one is #sin(pi/x) to 0#, because #pi/x to 0# when #x to +infty# and #sin(0)=0#. There are other (easier?) ways to solve this (see below) than the "brutal" application of L'Hôpital's rule, which is sometimes troublesome. Anyway, your question is precise so let's go ahead.
L'Hôpital's rule states the following: if #x_0 in mathbb{R} cup {pm infty}# and #f#,#g# are two functions such that #lim_{x to x_0} f(x)/g(x)# is an indeterminate form #0/0# or #pm infty/infty#, then only if the limit #lim_{x to x_0} {f'(x)} / {g'(x)}# exists, the following equality holds: #lim_{x to x_0} f(x) / g(x) =lim_{x to x_0} {f'(x)} / {g'(x)} #. Notice the specification in bold. That's the "troublesomeness" of the rule: if the limit of the ratio #[f'(x)}/{g'(x)}# doesn't exist, we can't conclude that the limit of #f(x)/g(x)# doesn't exist too. Here you can find an example of limit for which the rule isn't valid.
Back to our limit: we have to manipulate the expression so that the indeterminate form #+infty * 0# becomes #0/0# or #pm infty / infty#. How can a product be turned into a fraction? The following equality holds: #x=1/{1/x}# So we can use this in the former limit and get: #lim_{x to +infty} x sin (pi/x)=lim_{x to +infty} {sin (pi/x)}/{1/x}# Now #sin(pi/x) to 0# and #1/x to 0# as #x to +infty#, so we got the case #0/0#. Now we can compute derivatives and hope that the limit of their ratio exists: #f(x)=sin(pi/x) => f'(x)=-pi/x^2 cos(pi/x)# #g(x)=1/x => g'(x)=-1/x^2# #lim_{x to +infty} {-pi/x^2 cos(pi/x)}/{-1/x^2}=lim_{x to +infty} pi cos(pi/x)=pi *cos(0)=pi# The limit exists, so by L'Hôpital's rule we conclude that #lim_{x to +infty} x sin (pi/x)=pi#
Notice that this limit is easier to solve if you multiply the function by #1=(pi/x)/(pi/x)#: #lim_{x to +infty} x *(pi/x)/(pi/x) *sin (pi/x) =lim_{x to +infty} pi*sin(pi/x)/(pi/x)#. Now we can apply the following notable limit: #lim_{epsilon to 0} sin(epsilon) / epsilon = 1# Here we have that #epsilon=pi/x to 0# if #x to +infty#. So we get that #lim_{x to +infty} pi*sin(pi/x)/(pi/x)=pi*1=pi#
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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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