How to Evaluate Limits of a Function? Extraclass
HTML-код
- Опубликовано: 1 июн 2024
- How to evaluate limits of a function is a video which will solidify your concept of limits where you will learn to find limits by equating left hand limit to right hand limit.
JEE Advanced Course Kit
Launching on 1st Aug(Wednesday), 9:30 PM only on
JEE Under 1000.
Download Schedule : cutt.ly/UslBXz7
To Achieve JEE Advanced Rank Under 1000
Subscribe to this Channel
✅ / @jeeunder1000
✅Do subscribe and press the 🔔 bell icon to get notifications for all new videos / extraclass
How to evaluate limit of a function?
The first question that pops into
our mind is what is limit of a function?
We often come across functions which are not defined at particular values of x.
However sometimes the value of function as x approaches that particular number may symbolise a useful quantity.
For example, let’s say a car is moving in such a way that its displacement S is given by the relation S equal to t cube where t is the time taken.
Suppose we want to know that for such a relation, what will the instantaneous velocity be, say after 2 seconds.
So, the instantaneous velocity at t equal to 2 can be thought to be same as the velocity at t equal to 2 minus h and 2 plus h, where h is an infinitesimally small positive quantity approaching zero.
Thus the instantaneous velocity at t just less than 2 can be calculated as limit h tends to zero, 2 minus h cube minus 2 cube whole divided by minus h,
This value is known as the Left hand Limit as t approaches 2.
Similarly the instantaneous velocity at t just greater than 2 can be calculated.
This limit is known as the Right hand Limit.
Since LHL = RHL and is finite we can be sure of its existence at t equal 2.
So we can say that limit at t equal to 2 exists and the instantaneous velocity is 12 m/s.
Lets take an example to solidify our concept of limits.
Calculate limit of root of one minus cos 2 times x minus 1 whole divided by x minus 1 as x approaches 1. The options are.
This question was asked in IIT JEE 1998 exam
First let us see if we can simplify the given expression.
We know that cos 2 x is equal to
1 minus 2 times sine square x
So by simplifying we get limit x approaching 1, square root of 2 sin squared ‘x’ minus 1 divided by x minus 1.
Since constants don’t play a role in deciding the limiting value of a function we can consider the constant out of the limits.
Hence, our equation simplifies to root 2 limit mod sin x minus 1 upon x minus 1 as x approaches 1.
For Left Hand Limit lets put x equal to 1 minus h, where h is greater than zero, so for x approaching 1 minus, h approaches zero,
therefore, we get root 2 limit h approaching zero mod of sin minus h upon minus h.
Which is Equal to root 2 limit h approaching zero, sine h upon minus h which is equal to minus root 2.
Again, for Right Hand limit put x equal to 1 plus h where h is greater than zero.
Hence, for x approaching one plus, h approaches zero.
So, Right Hand Limit is Equal to limit h approaching zero root 2 mod sin h upon h.
Equal to limit h approaching zero root 2 sin h upon h which is equal to root 2.
#howtoevaluatelimitofafunction
#evaluationoflimits
#limits
Superb animation, loved the concept
great
Awesome animation , really helpful video
Glad you think so!
Awesome video loved the animation
Glad you enjoyed it
Please upload more videos.
Sure
Thanks Sir 👌👍........😍😍
You are welcome, do let us know what other videos you want us to post
PG Sir plz tell will sacred graphs come or not we dont have much time and sacred is a full series sir plz bring it
It's getting recorded, we will release it soon
@@Extraclass ohk. Sir we need it desparately for adv coz in adv there are many questions that are hard to solve via algebraic but can easily be done with graphs so thats why we need it as soon as possible