How to Evaluate Limits of a Function? Extraclass

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.
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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.
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