`lim_(x->0)(a^x-b^x)/x=log_e(a/b)`

lim_(x rarr0)(a^(x)-b^(x))/(x)=log_(e)((a)/(b))

Prove quad that quad (i) lim_(x rarr0)(a^(x)-1)/(x)=log_(e)aquad (ii) lim_(x rarr0)(log_(1+x))/(x)=1

lim_(x rarr0)((a^(x)-1)/(x))=log_(e)a

Using lim_(x rarr 0) (e^(x)-1)/(x)=1, deduce that, lim_(x rarr 0) (a^(x)-1)/(x)=log_(e)a [agt0].

(i) lim_(x to 0) (a^(x) - 1)/(log_(a)(1 + x)), a gt 0 (ii) lim__(x to 0) (In (X + a)- In a)/(e^(2x) - 1) (ii) lim_(x to (pi)/(4)) (In(tanx))/(1 - cotx)

(i) lim_(x to 0) (a^(x) - 1)/(log_(a)(1 + x)), a gt 0 (ii) lim__(x to 0) (In (X + a)- In a)/(e^(2x) - 1) (ii) lim_(x to (pi)/(4)) (In(tanx))/(1 - cotx)

Show that : lim_(x rarr0)((a^(x)-1)/(x))=log_(e)a

' lim_ (x to 0) (a^(x)-b^(x))/(e^(x)-1) is equal to

lim_(xrarr0) (a^x-b^x)/(e^x-1) is equal to