Hello maths friedns! (Q-F) hope everyone is fine ,can someone please help me solve this question?.I have tried it multiple times but always get stuck.Its question based on L hopital rule and i have provided answer as well.Would really appreciate the help.

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Answer 1 · 2018-03-23T16:36:04

#=>lim_(x->infty)(e^x+x)^(1/x)=e#

We begin with:

#lim_(x->infty)(e^x+x)^(1/x)#

We can simplify the power-to-a-power by using a logarithm.

#=lim_(x->infty)e^(ln((e^x+x)^(1/x)))#

We use another property of logarithms, where a power inside the argument can be brought out as a multiplicative factor.

#=color(blue)(lim_(x->infty)e^(1/xln(e^x+x))#

Now this is of the form #f(g(x))#. For limits, we can compute the limit #L# of #g(x)#. Then, we can rewrite our limit as follows:

#=>lim_(x->infty)f(g(x)) = lim_(u->L)f(u)#

Our #g(x)# is:

#g(x) = 1/x ln(e^x+x)#

The limit of #g(x)# is:

#lim_(x->infty)g(x) = lim_(x->infty) 1/xln(e^x+x)#

Using L'Hospital's rule:

#= lim_(x->infty) (1+e^x)/(e^x+x)#

Still no good. We use L'Hospital's rule a second time:

#=lim_(x->infty)e^x/e^x=color(red)(1)#

So the limit of #g(x)# is #1#.

Now we remember the limit chain rule we discussed before:

#=>lim_(x->infty)f(g(x)) = lim_(u->L)f(u)#

So we can rewrite our original limit as:

#color(blue)(lim_(x->infty)e^(1/xln(e^x+x))) = lim_(u->color(red)(1))e^(u)=e#

Hence:

#=>lim_(x->infty)(e^x+x)^(1/x)=e#