# Finding missing variables to make limit true.

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.

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

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.

• 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. – CalcLearn412 Apr 30 '18 at 0:58

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

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