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Answer 1 · 2017-02-15T23:37:33

#= 0#

#lim_(x to oo) x / 4^x#

There really isn't much to do here.

Consider these:

#2 / 4^2, 10 / 4^10, 100/ 4^100#

The exponential grows way more quickly than the numerator so the limit is zero.

However, we can persevere and say that, by definition: #4^x = e^((ln 4)^x) = e^(ln 4 * x) #

Thusly, by definition of #e^z#:

#4^x = 1 + (ln 4 * x) + ((ln 4 * x)^2)/(2!) + ... #

And so:

#lim_(x to oo) x / 4^x =lim_(x to oo) (x)/( 1 + (ln 4 * x) + ((ln 4 * x)^2)/(2!) + ... ) #

# =lim_(x to oo) (1)/( 1/x + (ln 4 ) + ((ln 4 * x))/(2!) + ... ) #

# = (1)/( lim_(x to oo) 1/x +lim_(x to oo) (ln 4 ) +lim_(x to oo) ((ln 4)^2 * x)/(2!) + lim_(x to oo) \O(x^2) ) #

# = (1)/( lim_(x to oo) \O(x) ) #

#= 0#