Question 361546
a.  Putting x = 0 produces the indeterminate form 0/0. But for this problem, there is NO need to use L'Hopital's rule as shown below:
But {{{ lim(x->0, (xcosx-sinx)/x)= lim(x->0, (xcosx/x))-lim(x->0, sinx/x)}}}
= {{{lim(x->0, (cosx))-lim(x->0, sinx/x)}}} = 1 -1 = 0.  

b.   Putting x = 0 produces the indeterminate form 0/0. 
Now{{{ lim(x->infinity, (e^x-2^x)/x )  =lim(x->infinity, (e^x-(ln2)*2^x) )  }}}, by direct application of L'Hopital's rule.  This does NOT produce an indeterminate form, so the limit is {{{e^0 - (ln2)*2^0 = 1-ln2}}}.

c.   {{{ lim(x->1,  (x/(x-1) - (1/lnx) ) ) }}},
 ={{{ lim(x->1, ( (xlnx-x+1)/((x-1)*lnx)) ) }}}, combining fractions.
={{{ lim(x->1, ( lnx/(lnx+(x-1)/x)) ) }}}, applying L-H rule. (Gives 0/0)
={{{ lim(x->1, ( (xlnx)/(xlnx+x-1)))}}}, simplifying complex fractions, (Gives 0/0)
={{{ lim(x->1, ( (lnx+1)/(lnx+2) ) )}}}, applying L-H rule (Not indeterminate anymore)
={{{1/2}}}