`lim_(x->0){loga+x-loga}/{x}=`

Evaluate: ("lim")_(x->a)(logx-loga)/(x-a)

Evaluate lim_(xtoa) (logx-loga)/(x-a).

Evaluate: ("lim")_(xveca)(logx-loga)/(x-a)

Evaluate: ("lim")_(xveca)(logx-loga)/(x-a)

lim_(x->0) (sin x /x)

lim_(x->0)(tanm x)/(tann x)

Let a = lim_(x->0) x cotx and b = lim_(x->0) xlog x, then

lim_(x->oo)cot^(-1)(x^(-a)log_a x)/(sec^(-1)(a^xlog_x a)),(a >1) is equal to (a) 2 (b) 1 (c) (log)_a2 (d) 0

lim_(x to 0) (log (1 + 2x))/(x) + lim_(x to 0) (x^(4) - 2^(4))/(x - 2) equals

lim_(x->0) ((1+x)^(1/x)-e)/x is equal to