How do I find the limit of a function as #x# approaches a number?

1 Answer
Oct 2, 2015

If the function is continuous at that value of #x#, just computing it should work. For other functions, try to factor the discontinuity out, and then computing it.

If nothing works, and you don't have access to derivatives to use L'Hôpital rule, just take a numeric approach and see if it's close to a pattern, or try to use other approaches like the Squeeze Theorem or a geomtrical proof. For example:

#lim_(x rarr4)(x^2-8x+16) = 0#

The function is continuous* for all real values of #x# so just plugging the value of #x# is good enough.

#lim_(xrarr4)(x^2-16)/(x-4) = lim_(xrarr4)((x+4)(x-4))/(x-4) = lim_(xrarr4)(x+4) = 8#

The function isn't continous at #x=4# because we'd have division by #0#, but using algebra we can take out the denominator, making a continuous function to just plug the number in.

#lim_(thetararr0)(sin(theta))/theta = 1#

This one you can't factor out. You need to use the unit circle and the Squeeze theorem to make a proof or try a numeric approach.

*A function is continuous on a certain range if you can draw the graph on that range without lifting your writing utensil, or more formally, if there isn't any point that would create a math error, like division by zero, even root of a negative, logarithm of a negative or null number, etc.