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Evaluate $displaystyle lim_mrightarrow inftybigg[m^3int^2m_mfracxdxx^5+1bigg]$ for $minmathbbN$

what i try

put $x^5+1=t$ and $dx=frac15x^-4dt=frac15x^4dt$

$$lim_nrightarrow inftyfracm^35bigg[int^32m^5+1_m^5+1fracx^2t(t-1)dtbigg]$$

How do i solve it. Help me please

We have $int^2m_mfracxdxx^5+1 leq mcdot frac2mm^5+1 stackrelmto inftylongrightarrow 0$

Set $I(m) = int^2m_mfracxdxx^5+1 Rightarrow m^3cdot I(m) = fracI(m)frac1m^3$.

So, we have a L'Hospital case of $frac00$:

begineqnarray* fracI(m)frac1m^3
& stackrelL'Hospt.sim & frac2frac2m(2m)^5+1-fracmm^5+1-3cdot m^-4 \
& = & -frac13left(frac4m^532m^5+1 - fracm^5m^5+1right) \
& stackrelm to inftylongrightarrow & frac724
endeqnarray*

Putting $x=mt$ in the integral we see that the desired limit is equal to the limit of the expression $$m^5int_1^2fract,dtm^5t^5+1$$ which is same as the limit $$lim_hto 0^+ int_1^2fract,dth+t^5$$ via the substitution $h=1/m^5$. Clearly the integrand is continuous in both $t, h$ and thus the limit can be taken inside to get the desired limit as $int_1^2t^-4,dt=7/24$.

Another simpler approach is to note that $$m^3int_m^2mfracdxx^4=frac724$$ and we have therefore $$left|m^3int_m^2mfracx,dx1+x^5-frac724right|=m^3int_m^2mfracdxx^4(1+x^5)leq m^3cdot mcdotfrac1m^4(1+m^5)$$ and the rightmost expression tends to $0$. This avoids the interchange of limit with integral and is fully within the scope of a typical high school calculus course.

Another intuive but non-rigorous argument. Intuitively as the lower bound of the integral in the denominator, $mto infty$ so the integrand goes to $0$ so we can say essentially the integral in the denominator goes to $0$. This is very inviting for an application of L'Hopital's rule which gives you $7/24$ as the limiting value as follows. We can approximate the integral by: $$beginalignedint_m^2mdfrac1x^4mathrm dx=dfrac-13(2m)^3+dfrac13m^3implies lim_mto inftym^3int_m^2mdfracxx^5+1mathrm dxto dfrac-124+dfrac13=boxeddfrac724endaligned$$