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Get Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 8 - Problem 4re
Get Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 8 - Problem 4re

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# Solution: Limits of sequences Evaluate the limit of the ISBN: 9780321570567 2

## Solution for problem 4RE Chapter 8

Calculus: Early Transcendentals | 1st Edition

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Problem 4RE

Limits of sequences Evaluate the limit of the sequence or state that it does not exist.

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STEP_BY_STEP SOLUTION Step-1 Exponential function; Exponential functions have the form f(x) = a x .Where ‘a’ is the base , and x is the exponent (or power) . If ‘a’ is greaterthan ‘1’ the function continuously increases in value as x increases. A specialproperty of exponential functions is that the slope of the function alsocontinuously increases as x increases. xExample ; the graph of y = 2 is ;\nNOTICE: As x increases , y also increases As x increases , the slope of the graph also increases.\n The curve passes through (0, 1). All exponential curves of the form f(x) =a x , passes through (0,1) , if a>0. The curve doesn’t passes through the x -axis .It just gets closer and closer to the x-axis as we take smaller and smaller x values. The natural exponential function y = e x is\n xRepresentation of exponential function is ; e Inverse is ln(x) x Derivative is e x Indefinite integral is e + C. x x x2 x3Exponential function expansion is e = 1 + 1!+ 2! + 3! +................... We , know the formula lim ( 1 + x )n = e …………………..(1) n n Step-2 3 2n The given sequence is a = (n1 + n ) Now , we have to find out the value of lim a n n 3 2n That is , lim a = n lim ( 1 + n ) n n = lim (( 1 + 3 ) ) . n n From (1) , we know that lim ( 1 + x )n = e x n n\nSo , the limit value becomes; lim (( 1 + 3 ) )2 = (e ) 2 = e . n n m n mnSince , (a ) = a . Therefore , lim a = n lim ( 1 + 3 )2n = e .6 n n n

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##### ISBN: 9780321570567

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