 Hello and welcome to the session. In this session, first we are going to discuss infinite limit of a function. Let f of x be a function of x if the value of x can be made greater than any preassigned number by taking x close to a, then the function f of x becomes positively infinite x approaches to a that is limit f of x as x tends to a is equal to less infinity. Similarly, if the value of f of x can be made less than any preassigned number by taking x close to a, then the function f of x becomes negatively infinite x approaches to a that is limit of f of x as x tends to a is equal to minus of infinity. Now we are going to discuss about trigonometric limits and first we have limit sin of alpha as alpha tends to 0 is equal to 0 and limit cos of alpha as alpha tends to 0 is equal to 1. Let alpha be an angle in the standard position. It is measured in radians, alpha is the angle between the lines rq and rp from the point p drop pq perpendicular to the x axis. Then we have sin of alpha which is given by perpendicular upon height continuous is equal to pq upon rp that is sin of alpha is equal to perpendicular pq upon height continuous op. When alpha is getting smaller and smaller then p will get closer and closer to q which implies that as alpha tends to 0 pq will tend to 0 that is pq by op will tend to 0. Therefore as alpha tends to 0 sin of alpha will also tend to 0. Now cos of alpha is given by perpendicular upon height continuous that is op upon op. Here we have cos alpha is equal to this op upon height continuous op is getting smaller and smaller then p will get closer and closer to q which implies that as alpha tends to 0 op will tend to op therefore op upon op will tend to 1. Therefore as alpha tends to 0 cos of alpha will tend to 1 op will tend to 1 and is equal to cos of alpha hence limit sin of alpha as alpha tends to 0 is equal to 0 and limit cos of alpha as alpha tends to 0 is equal to 1. Next we have limit sin of x by x as x tends to 0 is equal to 1 let there be a unit circle with center o this is a unit circle with center o let angle a o b is equal to x radians angle a o b is equal to x radians. Now we join a b and draw a c perpendicular to o a at a produce o b to meet a c at c. Now from this symbol it is clear that area of triangle o a b is less than area of sector o a b and is less than area of triangle o a c and the o a b is less than area of sector o a b is less than area of triangle o a c which implies that area of triangle o a b is equal to 1 by 2 into o a into o b into sin of x is less than area of sector o a b which is equal to 1 by 2 into o a square into x is less than area of triangle o a c which is given by 1 by 2 into o a into a c since we have taken a unit circle therefore the length of the radius o a is equal to 1 unit and the length of the radius o b is also equal to 1 unit. Therefore we have 1 by 2 into o a that is 1 into o b that is 1 into sin of x is less than 1 by 2 into o a square that is 1 square into x is less than 1 by 2 into o a that is 1 into a c which is equal to sin of x a c is the right-hand triangle therefore sin of x is given by the perpendicular a c by this o a so we have a c is equal to o a into sin of x and we know that o a is equal to 1 so we have 1 into sin of x which is equal to sin of x so a c is equal to sin of x therefore we have 1 by 2 into sin of x is less than 1 by 2 into x is less than 1 by 2 into sin of x which can be written as sin of x is less than x is less than tan of x which implies that 1 is less than x by sin of x is less than 1 by cross of x and we have 1 is greater than sin of x by x is greater than cross of x therefore the value of sin of x by x lies in between 1 and cross of x from a previous read write we know that limit cross of x as x tends to 0 is equal to 1 therefore when x tends to 0 the value of sin of x by x lies in between 1 and 1 that is limit sin of x as x tends to 0 is equal to 1 or we can also write it as limit x by sin of x as x tends to 0 is equal to 1 here x is expressed in radians and we should note that if x is given in degrees that is x degrees then it should be converted into radians by using the formula if degrees is equal to pi into x upon 180 radians and then we have limit tan of x by x as x tends to 0 is equal to 1 we have limit tan of x by x as x tends to 0 which can be written as limit tan of x is equal to sin of x by cross of x whole upon x as x tends to 0 which can also be written as limit sin of x by x into 1 upon cross of x as x tends to 0 that is limit sin of x by x tends to 0 into limit 1 upon cross of x as x tends to 0 in a previous read write we know that limit sin of x by x as x tends to 0 is equal to 1 into limit 1 upon cross of x as x tends to 0 so we have 1 into now putting the value of x as 0 in this expression we get 1 upon cross of 0 which is equal to 1 into 1 upon cross of 0 that is 1 upon 1 as cross of 0 is equal to 1 so we have 1 into 1 that is 1 hence limit tan of x by x as x tends to 0 is equal to 1 now we shall discuss some special limits first we have limit e raise to power x minus 1 whole upon x as x tends to 0 is equal to 1 next we have limit log of 1 plus x by x as x tends to 0 is equal to 1 then we have limit a raise to power x minus 1 upon x as x tends to 0 is equal to log of a to the base e next is limit 1 plus 1 upon x whole raise to power x as x tends to infinity is equal to e then we have limit 1 plus x whole raise to power 1 upon x as x tends to 0 is equal to e next is limit 1 plus a upon x whole raise to power x as x tends to infinity is equal to e raise to power a now we shall discuss in the terminate limits if f of x takes any of the form that is 0 by 0 infinity by infinity infinity minus infinity 0 into infinity 0 raise to power 0 1 raise to power infinity and infinity raise to power 0 at x is equal to a then f of x is indeterminate at x is equal to a now we are going to discuss l-hopper's rule l-hopper's rule states that if f of a by g of a is of the form 0 by 0 or infinity by infinity then limit f of x by g of x as x tends to a is equal to limit f dash of x by g dash of x as x tends to a that is differentiating numerator and denominator separately now we shall learn about geometrical interpretation of dy by dx let us consider the curve y is equal to f of x let f of x be a differentiable function let p with the coordinates x f of x and q with the coordinates x plus delta x f of x plus delta x be two points on the curve from the points p and q draw pl and qm perpendicular to the x axis pn be perpendicular to nq then slope of t called pq is given by tan of the angle qtn since we know that qtn is the right angle triangle therefore tan of angle qtn is given by perpendicular upon base that is qn upon pn therefore we can write tan of angle qtn is equal to qn upon pn and qn is given by f of x plus delta x minus f of x and pn is equal to delta x therefore we have qn that is f of x plus delta x minus f of x upon pn which is given by delta x let q tends to p that is we have delta x tends to zero so we have limit slope of called pq q tends to p is equal to limit f of x plus delta x minus f of x whole upon delta x as delta x tends to zero as q tends to p called pq tends to the tangent to y is equal to f of x at p therefore slope of the tangent at p is equal to limit f of x plus delta x minus f of x by delta x as delta x tends to zero which implies that slope of the tangent at p is equal to limit f of x plus delta x minus f of x upon delta x as delta x tends to zero which can be written as f dash of x that is we have tan of psi is equal to f dash of x where psi is the inclination of the tangent to the curve y is equal to f of x at the point p with coordinates x f of x with the axis therefore we have dy by dx is equal to tan of psi hence the derivative at any point p is the tangent of the angle that the tangent to the curve at p makes with the positive direction of x axis and thus follows the two results first is if the tangent is parallel to the x axis then psi is equal to zero therefore dy by dx is equal to tan of psi which is equal to tan of zero that is zero and if the tangent is perpendicular to the x axis then psi is equal to 90 degrees and cot of psi that is cot of 90 degrees is equal to zero therefore we have dx by dy is equal to zero or we can also write it as dy by dx is equal to infinity this completes our session hope you enjoyed this session