How do you find #\lim _ { x \rightarrow + \infty } \frac { x } { \ln ( 1+ 2e ^ { x } ) }#?

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Answer 1 · 2017-09-24T16:25:47

Because the expression evaluated at #oo# results in the indeterminate form, #oo/oo#, one should use L'Hôpital's rule

Given: #lim_(x to +oo) x/ln ( 1+ 2e^x)#

Differentiate both the numerator and denominator:

#lim_(xto+oo) ((d(x))/dx)/((d(ln ( 1+ 2e^x)))/dx) = lim_(xto+oo) 1/((2e^x)/( 1+ 2e^x))#

Division by a fraction is the same as multiplication by the reciprocal of the denominator:

#lim_(xto+oo) ( 1+ 2e^x)/(2e^x)#

Multiply the by 1 in the form #e^-x/e^-x#

#lim_(xto+oo) ( e^-x+ 2)/2 = (0+2)/2 = 1#

L'Hôpital's rule states that as this limit goes so goes the original limit:

#lim_(x to +oo) x/ln ( 1+ 2e^x) = 1#