# FORMULAS IN EVALUATING LIMITS

(1)  lim θ -> 0 sin θ / θ  =  1

(2)  lim θ -> 0 (1 - cos θ) / θ  =  0

(3)  lim θ -> 0 tan θ / θ  =  1

(4)  lim x -> 0 (ex - 1) /x  =  1

(5)  lim x -> 0 (ax - 1) /x  =  log a

(6)  lim x -> 0 log (1 + x)/x  =  1

(7)  lim x -> 0 sin-1x/x  =  1

(8) lim x -> 0 tan-1x/x  =  1

(9) lim x ->  (1 + 1/x)x exists and this limit is e.

(10) lim x ->0 (1 + x)1/x   =  e

(11)  lim x ->  (1 + k/x)x  =  ek

(12)  lim x -> a (xn - an)/(x - a)  =  nan-1

(13)  lim x -> a sin (x - a)/(x - a)  =  1

(14)  lim x -> a tan (x - a)/(x - a)  =  1

This number e is also known as transcendental number in the sense that e never satisfies a polynomial (algebraic) equation of the form

a0xn + a1xn-1 + ............. + an-1 x + an  =  0

## Practice Questions

Question 1 :

Evaluate the following limit

lim  x -> (1 + 1/x)7x

Solution  :

=  lim  x -> ∞ (1 + 1/x)7x

The given question exactly matches the formula

lim x ->  (1 + 1/x)x  =  e

lim  x -> ∞ (1 + 1/x)7x  =  lim  x -> ∞ ((1 + 1/x)x)7

=  e7

Hence the value of lim  x -> ∞ (1 + 1/x)7x is e7.

Question 2 :

Evaluate the following limit

lim  x -> 0 (1 + x)1/3x

Solution  :

=  lim  x -> 0 (1 + x)1/3x

The given question exactly matches the formula

lim x ->0 (1 + x)1/x   =  e

lim  x -> 0 (1 + x)1/3x  =  lim  x -> 0 ((1 + x)1/x)1/3

=  e1/3

Hence the value of lim  x -> 0 (1 + x)1/3x is e1/3.

Question 3 :

Evaluate the following limit

lim  x ->  (1 + k/x)m/x

Solution  :

=  lim  x ->  (1 + k/x)m/x

The given question does not matches any of the formula. So let us apply the given limit directly in the question.

lim  x ->  (1 + k/x)m/x   (1 + k/)m/

(1 + 0)0

=  1

Hence the value of lim  x ->  (1 + k/x)m/x is 1.

Question 4 :

Evaluate the following limit

lim  x ->  [(2x2 + 3)/(2x2 + 5)]^(8x2 + 3)

Solution  :

=limx ->[(2x2+3)/(2x2+5)]^8x2 limx->[(2x2+3)/(2x2+5)]3

Part 1 :

=  limx->([(1+3/2x2)/(1+5/2x)]^2x2)4

By distributing the limit to the numerator and denominator, we get

=   limx->([(1+3/2x2)^2x2)4 /limx->([(1+5/2x2)^2x2)4

This exactly matches the formula limx->(1 + k/x)x  =  ek

=   (e3)4 / (e5)4

=   e12/ e20

=  1/e8

Part 2 :

=  limx->[(2x2+3)/(2x2+5)]3

=  limx->[(1+3/2x2)/(1+5/2x2)]3

=  limx->[(1+3/2x2)]3/limx->[(1+5/2x2)]3

=  1

=  1/e8 (1)  =  1/e8

Question 5 :

Evaluate the following limit

lim  x ->  (1 + (3/x)) x + 2

Solution  :

=  lim  x ->  (1 + (3/x)) x + 2

=  lim  x ->  (1 + (3/x))x  ⋅ lim  x ->  (1 + (3/x))2

=  e ⋅ lim  x ->  (1 + (3/x))2

By applying the limit value, we get

=  e ⋅ (1 + 0)2

=  e

Hence the value of lim  x ->  (1 + (3/x)) x + 2 is e. Apart from the stuff given in this section, if you need any other stuff in math, please use our google custom search here.

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