lim_x -> 0 (48 / x^4) integral _0^x (t^3 / t^6 + 1) dt is equal to ___________.

48 limlimits_x -> 0 fracintegral _0^x (t^3 / t^6+1) d tx^4((0 / 0)) Applying L' Hospitals Rule 48 lim _x -> 0 (x^3 / x^6+1) x (1 / 4 x^3) \ = 48 / 4mathop lim limits_x -> 0 1 / x^6 + 1 = 12

limx arrow 0 frac 48x4 t0x frac t3t61 d t is equal to | null (Mathematics)

48 \lim\limits_{x \rightarrow 0} \frac{\int_0^x \frac{t^3}{t^6+1} d t}{x^4}\left(\frac{0}{0}\right)

Applying L' Hospitals Rule

\begin{aligned} 48 \lim _{x \rightarrow 0} \frac{x^3}{x^6+1} \times \frac{1}{4 x^3}\\\\ \end{aligned}

{{48} \over 4}

\mathop {\lim }\limits_{x \to 0} {{1} \over {{x^6} + 1}} = 12

Explanation

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