The key concept to use in this question is l'hopital's rule.
Firstly, when you directly substitute x by 0, the limit looks like this: ( tan(0) ) * ( ln(0) ). As tan(0) equals zero and ln(0) equals negative infinity, this limit is in the form of 0 times negative infinity, which is an indeterminate form. Hence, we are able to use the l'hopital's rule. However, we need to convert this limit into a some kind of fraction to use this rule.
Hence, I would change the limit like this: lim (x --> 0+) ( ln4x / cot7x ).
Now, use the l'hopital's rule and take the derivative of both the numerator and the denominator:
Hi James,
Thank you for your question.
The key concept to use in this question is l'hopital's rule.
Firstly, when you directly substitute x by 0, the limit looks like this: ( tan(0) ) * ( ln(0) ). As tan(0) equals zero and ln(0) equals negative infinity, this limit is in the form of 0 times negative infinity, which is an indeterminate form. Hence, we are able to use the l'hopital's rule. However, we need to convert this limit into a some kind of fraction to use this rule.
Hence, I would change the limit like this: lim (x --> 0+) ( ln4x / cot7x ).
Now, use the l'hopital's rule and take the derivative of both the numerator and the denominator:
lim (x --> 0+) ( ln4x / cot7x ). = lim (x --> 0+) ( (4/4x) / (-7(csc7x)^2) ) = lim (x --> 0+) ( (1/x) / (-7(csc7x)^2) )
= lim (x --> 0+) ( ((sin7x)^2) / -7x ).
When you substitute x with 0 in this new limit, you get 0 / 0, which is another indeterminate form. Hence, you can use the l'hopital's rule again.
lim (x --> 0+) ( (-7(sin7x)^2) / x ) = lim (x --> 0+) ( (14(sin7x)(cos7x)) / -7 ) = lim (x --> 0+) (-2sin(7x)cos(7x)
If you substitue x with 0 to this new limit, you get an answer of 0.
Hence, the value of the limit is 0.
Does this answer your question?