Good evening class, today we will talk about Limits. Let us talk about few simple formulae of limits,

limit extending to a, x raised to par n minus a raised to par n upon x minus a is equal to n into a2 par

n minus 1. Limit extending to zero, e raised to par x minus 1 upon x is equal to 1. Put it as x to zero

and it is zero by zero format and the limiting value comes as 1. Similarly limit extending to zero, a

raised to par x minus 1 upon x is log a to the base e.

More formulae on limits, Limit extending to zero, log of 1 plus x upon x is 1. Limit extending to zero,

log of 1 plus x upon x, log to the base a is 1 upon log a to the base e.

Limit extending to zero sin x upon x is also 1. Limit extending to zero tan x upon x is also 1. Similarly

limit extending to zero sin inverse x upon x is also 1. Limit extending to zero tan inverse x upon x is

also 1. Good.

But how do we use this formulae, we use this formulae as limit extending to zero sin 2x upon x. Say

we have to find this. I will multiply it and divide it by 2. Sin zero upon same zero or what we can do is

we can put 2x equals to t limit t extends to zero, as x extends to zero, t also extends to zero. Sin t

upon t is 2 into 1 that is 2. Limit t extending to zero is sin t upon t is 1.

Next let us talk about L-Hospital’s rule, L-Hospital’s rule. It is applicable for the limits of the form zero

by zero or infinity by infinity. It is applicable for the limits of the form zero by zero or infinity by

infinity. It says that limit extending to a, fx upon gx is equal to limit extending to a f dash x upon g

dash x. Differentiate the above function separately and differentiate the below function separately.

And you get the value of limit.

For example, zero by zero format, differentiating numerator we get n into x to par n minus 1

denominator we get 1. Now put x equals to a, we will get the value of limit that is n into a to par n

minus 1.

One more question of L-Hospital say, find limit extending to zero, x minus sin x upon xq. Applying L-

Hospital rule we get differentiate the numerator 1 minus cos x denominator is 3x square. Format is

zero by zero, differentiate above separately and below separately. Put x as zero and see. Cos zero, 1

minus 1 zero, now also zero by zero format, and again L-Hospital’s rule it becomes, and limit

extending to zero, sin x upon x put 1, the answer becomes 1 by 6. Simple application of L-Hospital’s

rule.

Let us talk about questions, first question of your assignment. Put x as 1 and format is zero by zero.

Applying L-Hospital’s rule we get, now put, it is extending to 1, now put x to 1 we get 8 minus 2 upon

4 minus 2 that is 6 by 2 that is 3. Answer of this question is 3. Simple it was a zero by zero format.

We have applied L-Hospital’s rule, differentiate above separately and below separately and we got

the answer.

Second question of your assignment, put x as 1 and see, zero by zero format. So this limit L is equal

to, below there are factors 6 minus 1 into 3x plus 7, x minus 1 into 3x plus 7. x minus was again

factorized.