Infinite Limits

追加: 2026-06-02

[00:00:00]so note that there are certain cases wherein l is a non-zero constant value while our denominator is zero so the question is when is it correct to say that a limit does not exist and when is it correct to say that a limit is positive infinity or negative infinity so here we are given a theorem but for us to understand this theorem more we will be looking at a short illustration all right so let's consider g of x is equal to 5x all over 4 minus x squared and we want to get the limit so we want to determine the limit of g of x as x approaches 2.

[00:00:52]so by a direct substitution you can see here that the numerator is 10 while the denominator is zero okay so um usually we would just write this as a does not exist but here we want to deter we want to use a different approach and so we want to evaluate the limit by analyzing separately the behavior of the numerator and the denominator as x approaches 2 from both the left and the right okay so let's start with the limit of g of x as x approaches 2 from the left so when we substitute any x value that is less than 2 let's say 1.99 so first we have here the limit of 5x all over 4 minus x squared as x approaches 2 from the left so for our numerator if we substitute any x value that is less than 2 then the numerator can be seen to be approximately or can be seen to be very close to 10 right so the numerator would be 10 and the denominator so if we plug in 2 is close to zero however note that the denominator should be a relatively small number a relatively small positive number since four minus x squared is greater than zero for the values of x less than 2.

[00:02:37]since here we are looking at the limit of 4 minus x squared as x approaches 2 from the left so values of x less than 2.

[00:02:46]so we denote this as 0 from the red okay since uh zero since uh the denominator is a positive number okay does this means that the function increases without bound and is positive because both the numerator and the denominator are positive so through number one of the theorem as you can see here since l is greater than zero or the numerator is greater than zero and the denominator approaches zero through positive values then we can see that the limit limit of 5x all over 4 minus x squared as x approaches 2 from the left is equal to positive infinity okay so that's for the limit of g of x as x approaches 2 from the left so now we want to look at the limit of g of x as x approaches 2 from the right okay so here again we express this as the limit of 5 x all over 4 minus x squared as x approaches 2 from the right so here when we substitute any x value that is greater than 2 let's say 2.01 then the numerator is closer to 10 and the denominator is a negative number that is close to zero so now then the denominator would be a relatively small number that is negative so does we uh we express this as zero through negative values okay and that's because the denominator um well or uh the denominator is negative because uh for values of x that are greater than 2 4 minus x squared will always be a negative number okay does this means that the function decreases without bound and is negative because the numerator is positive while the denominator is negative so through number two of the theorem since our l is greater than zero and our denominator approaches 0 through negative values does we write that the limit here so the limit of 5x all over 4 minus x squared as x approaches 2 from the right is equal to negative infinity okay then now that we have the limits of g of x from the left and from the right so since they're not equal this means that the limit of g of x as x approaches 2 is now or does not exist so this means that because uh from the left the limit would approach positive infinity well from the right the limit would approach negative infinity does the limit of g of x as x approaches 2 itself is now or does not exist so this is now our final answer here