How do you find the limit of # (5^t-3^t)/t# as t approaches 0?

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Answer 1 · 2018-06-27T19:05:59

# ln(5/3)#.

Prerequisite : #lim_(h to 0)(a^h-1)/h=lna, (a gt 0)#.

#"The Limit"=lim_(t to 0)(5^t-3^t)/t#,

#=lim_(t to 0){(5^t-1)-(3^t-1)}/t#

#=lim_(t to 0){(5^t-1)/t-(3^t-1)/t}#,

#=ln5-ln3#,

#=ln(5/3)#.

Answer 2 · 2018-06-27T19:13:14

#lim_(t to0)(5^t-3^t)/t=ln(5/3)#

We know that,

#color(red)((1)lim_(hto0)(a^h-1)/h=lna ,where, ain RR^+ -{1}#

Let ,

#L=lim_(t to0)(5^t-3^t)/t#

#L=lim_(t to0)[(5^tcolor(blue)(-1)-3^tcolor(blue)(+1))/t]#

#L=lim_(t to0)(5^t-1)/t-lim_(t to0)(3^t-1)/t...tocolor(red)(Apply(1)#

#L=ln5-ln3to[Use:lnA-lnB=ln(A/B)]#

#L=ln(5/3)#