evaluate lim h approaches 0 (2+h)^4-16/h?

Question

let f(x) = x^2/x^3-5x^2  if x=/ 0, let f(0) =0. Classify the discontinuities of f(if any) as either remova ble, jump or infinite.

let f(x)={x^2      x<- 0
            {x^3+1  x>0   Determine if f is continuous at 0.  If not, classify the disconuity as removable, jump or infinite.

Hemant · Accepted Answer

L = lim (h→0) [ ( 2 + h )⁴ - 16 ] / h

= lim ( h → 0) [ ( 2 + h )⁴ - 2⁴ ] / h

= lim ( h → 0) [ ( 2 + h )² + 2² ]·[ ( 2 + h )² - 2² ] / h

= [ ( 2 + 0 )² + 2² ] • lim (h→0) [ ( 2 + h ) - 2 ]·[ (2 + h ) + 2 ] / h

= [ 4 + 4 ] • lim (h→0) [ ( 2 + h ) + 2 ]

= [ 8 ] • [ ( 2 + 0 ) + 2 ]

= 32 ....................................... Ans.
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Happy To Help !
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havey · Answer

think of to apply L'Hopital's Rule Lim f(x) / g(x) = Lim f'(x) / g'(x) x ? a?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ???? ?? ?? ?? ?? ????x ? a Lim [?(a million + h) - a million] / h = ? x ? 0 f(h) = ?(a million + h) - a million = [(a million + h)^(a million/2)] - a million It appears like: [u^(a million/2)] - a million ? the place: u = (a million + h) ? u' = a million The derivate is: (a million/2) * u' * u^[(a million/2) - a million] f'(h) = (a million/2) * a million * (a million + h)^[(a million/2) - a million] f'(h) = (a million/2) * (a million + h)^(- a million/2) f'(h) = (a million/2)/(a million + h)^(a million/2) f'(h) = (a million/2)/?(a million + h) f'(h) = a million/2?(a million + h) g(h) = h g'(h) = a million f'(h) / g'(h) = [a million/2?(a million + h)] / a million f'(h) / g'(h) = a million/2?(a million + h) while h procedures 0 ? a million/2?(a million + h) = a million/2?(a million + 0) = a million/2?a million = a million/2 end: Lim [?(a million + h) - a million] / h = a million/2 x ? 0

? · Answer

think of to apply L'Hopital's Rule Lim f(x) / g(x) = Lim f'(x) / g'(x) x ? a?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ???? ?? ?? ?? ?? ????x ? a Lim [?(one million + h) - one million] / h = ? x ? 0 f(h) = ?(one million + h) - one million = [(one million + h)^(one million/2)] - one million It sounds like: [u^(one million/2)] - one million ? the place: u = (one million + h) ? u' = one million The derivate is: (one million/2) * u' * u^[(one million/2) - one million] f'(h) = (one million/2) * one million * (one million + h)^[(one million/2) - one million] f'(h) = (one million/2) * (one million + h)^(- one million/2) f'(h) = (one million/2)/(one million + h)^(one million/2) f'(h) = (one million/2)/?(one million + h) f'(h) = one million/2?(one million + h) g(h) = h g'(h) = one million f'(h) / g'(h) = [one million/2?(one million + h)] / one million f'(h) / g'(h) = one million/2?(one million + h) while h strategies 0 ? one million/2?(one million + h) = one million/2?(one million + 0) = one million/2?one million = one million/2 end: Lim [?(one million + h) - one million] / h = one million/2 x ? 0

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evaluate lim h approaches 0 (2+h)^4-16/h?

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