text: # Video: Finding Limits Involving Trigonometric Functions
  
  Find lim_(𝑥 → 0) (sin² 7𝑥 + 3 tan² 3𝑥)/(8𝑥²).
  
  01:28
  
  ### Video Transcript
  
  Find the limit as 𝑥 approaches zero of sin squared of seven 𝑥 plus three tan squared of three 𝑥 over eight 𝑥 squared.
  
  If we were to try direct substitution, we would obtain zero over zero, which is undefined. Let’s try to find this limit using these rules. We’ll also be using the fact that the limit of a sum of functions is equal to the sum of the limits of the functions. Hence, we can write our limit as the sum of these two limits. We notice that we can take a factor of one-eighth out of the first limit and a factor of three-eighths out of the second limit.
  
  Next, we notice that both numerators and both denominators are squares, enabling us to write our limits like this. Now we can use the fact that the limit of a square of a function is equal to the square of the limit of the function. In doing this, we are left with this. And we notice that our limits look very similar to the ones we wrote out at the start.
  
  Substituting 𝑎 equals seven into the first limit rule, we see that our limit on the left must be equal to seven. And substituting 𝑎 equals three into the second limit rule, we see that our limit on the right must be equal to three. We obtain one-eighth multiplied by seven squared plus three-eighths multiplied by three squared. Simplifying this down, we obtain a solution of 19 over two.