0
$\begingroup$

I've been given a question for my calculus class which is:

$$\lim_{x\to 0}\left(\frac{\tan(2x)}{x^3}+\frac{a}{x^2}+\frac{\sin(bx)}{x}\right)=0$$

For what values of a and b is the following limit true?

I understand that you have to apply L'Hôpital's rule to solve it but I can't get my head around this particular quesiton.

$\endgroup$
1
  • 1
    $\begingroup$ Did you try setting up a common denominator before applying L'Hospital? What did you get after the first application? $\endgroup$ – user296602 Apr 30 '18 at 0:41
0
$\begingroup$

Here is a start: After writing a common denominator and applying L'Hospital's rule once, you should have the requirement

$$\lim_{x \to 0}\frac{2 \sec^2 2x + a + 2x \sin bx + b x^2 \cos bx}{x^2} = 0.$$

Rewrite this as

$$\lim_{x \to 0} \frac{2 \sec^2 2x + a}{x^2} + 2b + b = 0.$$

This should give you enough to compute $a$ and $b$.


Alternative solution: We have

$$b + \lim_{x \to 0} \frac{\tan(2x) + ax}{x^3} = 0.$$ Now $\tan (2x) = 2x + 8x^3 / 3 + O(x^5)$ via Taylor series, so $a = -2$ again.

$\endgroup$
1
  • $\begingroup$ That definitely clarifies things a bit. My lecturer's notes didn't explain the concept of L'Hopital's rule for these kinds of questions. $\endgroup$ – CalcLearn412 Apr 30 '18 at 0:58
0
$\begingroup$

Hint: Use the development of $tan(2x)= 2x+{1\over 3}(2x)^3+O(x^3)$

$sin(bx)= bx-{1\over 6}(bx)^3+O(x^3)$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.