So you plug the number in, but what if it was going to different numbers from both sides and the the left and right handed limits aren’t equal would we still just plug it in?
The only time a one-sided limit will not exist is if you have division by zero or if you are trying to take the even root of a negative number, such as a square root. For example, in the problem I did in this video with sqrt(16-x^2), the limit as we approach 4 from the positive side will not exist. This is because any number larger than 4 will cause a negative number under the square root. The domain for this problem is not defined for numbers larger than 4, which is why the righthand limit does not exist. For our second example, we were able to use algebraic methods to avoid division by zero, so this limit exists. However, if we are trying to find the one-sided limit at -2, this would not exist since we would still have division by zero even after factoring and cancelling. Therefore, if you do not have division by zero or the even root of a negative number, then the one-sided limit will exist. To see if a regular limit exist (not one-sided), the left and right hand limits must both exist at that number.
@@hutchmath Thanks a lot sir! I am learning calculus by my own!! Was having problem with this thing! Thanks a lot for such a clear explanation! Pray for me so that I can complete calculus to start with calculus based physics
ok but the thing that I don't understand when solving one sided limits algebraically is the following: You can do the same thing for lim x->2- and get -(1/4) as well ?? also what if the limit for a function when x->a has different values for a-->(a-) and x-->(a+) ?
Yes, the lim x->2- is also -1/4. The limit from the left and right approach the same number on the included graph. For your second question, it it were different values this means they would have to give you either the graph or a piecewise function in order to determine the limit.
@@hutchmath So in general, one-sided limits may be different for piecewise functions, but for most other functions, the left-hand and right-hand limits are the same?
it should still equal 0 since all the numbers after 4 are in the argand plane, so if you get closer and closer to 4 from the right side you would still end up with 0
@@whatislife233 yes, but you can graph it on the imaginary plane and technically there would be a right side. but i guess the limit from the right would not exist in the normal cartesian plane
The algebraic processes will be the same. The only time there is a difference is if you have division by zero and there is nothing you can do algebraically to prevent division by zero. In this case the limit will go to either infinity or negative infinity.
@@hutchmath Thanks for writing back. Even in that case, the algebraic calculations will be the same in the case of of one-sided vs. normal (from both sides) limits. It is simply that in order to see if a normal limit is approaching infinity, we would calculate both + and - and see if they both match. Please correct me if I am wrong, thanks again!
what a king, explained it so clearly
Glad my video helped you!
Wow in 5 minutes I understand more about it than multiple hour long lectures
this video is much better than my lecturer.
Thank you sir much respect🙏❤️
So you plug the number in, but what if it was going to different numbers from both sides and the the left and right handed limits aren’t equal would we still just plug it in?
But what if the limit doesnt exist? How can I tell it by algebric method? They perhaps return the same value😅..plz explain ne:(
The only time a one-sided limit will not exist is if you have division by zero or if you are trying to take the even root of a negative number, such as a square root. For example, in the problem I did in this video with sqrt(16-x^2), the limit as we approach 4 from the positive side will not exist. This is because any number larger than 4 will cause a negative number under the square root. The domain for this problem is not defined for numbers larger than 4, which is why the righthand limit does not exist. For our second example, we were able to use algebraic methods to avoid division by zero, so this limit exists. However, if we are trying to find the one-sided limit at -2, this would not exist since we would still have division by zero even after factoring and cancelling. Therefore, if you do not have division by zero or the even root of a negative number, then the one-sided limit will exist. To see if a regular limit exist (not one-sided), the left and right hand limits must both exist at that number.
@@hutchmath Thanks a lot sir! I am learning calculus by my own!! Was having problem with this thing! Thanks a lot for such a clear explanation! Pray for me so that I can complete calculus to start with calculus based physics
ok but the thing that I don't understand when solving one sided limits algebraically is the following: You can do the same thing for lim x->2- and get -(1/4) as well ??
also what if the limit for a function when x->a has different values for a-->(a-) and x-->(a+) ?
Yes, the lim x->2- is also -1/4. The limit from the left and right approach the same number on the included graph. For your second question, it it were different values this means they would have to give you either the graph or a piecewise function in order to determine the limit.
@@hutchmath So in general, one-sided limits may be different for piecewise functions, but for most other functions, the left-hand and right-hand limits are the same?
So in the first problem what if you approached 4 from the right, it would be DNE right? Thank you!
it should still equal 0 since all the numbers after 4 are in the argand plane, so if you get closer and closer to 4 from the right side you would still end up with 0
@@GiveMeThatSwordPower there is no right side tho
@@whatislife233 yes, but you can graph it on the imaginary plane and technically there would be a right side. but i guess the limit from the right would not exist in the normal cartesian plane
So there is no difference in calculating one sided limits vs. calculating normal limits?
The algebraic processes will be the same. The only time there is a difference is if you have division by zero and there is nothing you can do algebraically to prevent division by zero. In this case the limit will go to either infinity or negative infinity.
@@hutchmath Thanks for writing back. Even in that case, the algebraic calculations will be the same in the case of of one-sided vs. normal (from both sides) limits. It is simply that in order to see if a normal limit is approaching infinity, we would calculate both + and - and see if they both match. Please correct me if I am wrong, thanks again!
thank you about to go to my final huge help
So sir,I can simplify for reach one sided limit if needed?
you used my problem as an example how tf did u know
They always know
شكرا ياخي ♥️♥️
u saved my ass, thank you so so much