Notes: `lim_(x->oo) x^3/e^(x/2)` Evaluate the limit, using L’Hôpital’s Rule if necessary.

Tuesday, 14 February 2017

$\displaystyle\lim_{{{x}\to\infty}}\frac{{{x}}^{{3}}}{{{e}}^{{\frac{{x}}{{2}}}}}$ Evaluate the limit, using L’Hôpital’s Rule if necessary.

Given to solve,

$\displaystyle\lim_{{{x}\to\infty}}\frac{{{x}}^{{3}}}{{{{e}}^{{\frac{{x}}{{2}}}}}}$

as $\displaystyle{x}\to\infty$ then the $\displaystyle\frac{{{x}}^{{3}}}{{{{e}}^{{\frac{{x}}{{2}}}}}}=\frac{\infty}{\infty}$ form

so upon applying the L 'Hopital rule we get the solution as follows,

as for the general equation it is as follows

$\displaystyle\lim_{{{x}\to{a}}}\frac{{f{{\left({x}\right)}}}}{{g{{\left({x}\right)}}}}{i}{s}=\frac{{0}}{{0}}$ or $\displaystyle\frac{{\pm\infty}}{{\pm\infty}}$ then by using the L'Hopital Rule we get the solution with the below form.

$\displaystyle\lim_{{{x}\to{a}}}\frac{{{f{'}}{\left({x}\right)}}}{{{g{'}}{\left({x}\right)}}}$

so , now evaluating

$\displaystyle\lim_{{{x}\to\infty}}\frac{{{{x}}^{{3}}}}{{{{e}}^{{\frac{{x}}{{2}}}}}}$

= $\displaystyle\lim_{{{x}\to\infty}}\frac{{{\left({{x}}^{{3}}\right)}'}}{{{\left({{e}}^{{\frac{{x}}{{2}}}}\right)}'}}$

= $\displaystyle\lim_{{{x}\to\infty}}\frac{{{3}{{x}}^{{2}}}}{{{\left({{e}}^{{\frac{{x}}{{2}}}}\right)}{\left(\frac{{1}}{{2}}\right)}}}$

again $\displaystyle\frac{{{3}{{x}}^{{2}}}}{{{\left({{e}}^{{\frac{{x}}{{2}}}}\right)}{\left(\frac{{1}}{{2}}\right)}}}$ is of the form $\displaystyle\frac{\infty}{\infty}$ ...

Given to solve,

$\displaystyle\lim_{{{x}\to\infty}}\frac{{{x}}^{{3}}}{{{{e}}^{{\frac{{x}}{{2}}}}}}$

as $\displaystyle{x}\to\infty$ then the $\displaystyle\frac{{{x}}^{{3}}}{{{{e}}^{{\frac{{x}}{{2}}}}}}=\frac{\infty}{\infty}$ form

so upon applying the L 'Hopital rule we get the solution as follows,

as for the general equation it is as follows

$\displaystyle\lim_{{{x}\to{a}}}\frac{{{f{'}}{\left({x}\right)}}}{{{g{'}}{\left({x}\right)}}}$

so , now evaluating

$\displaystyle\lim_{{{x}\to\infty}}\frac{{{{x}}^{{3}}}}{{{{e}}^{{\frac{{x}}{{2}}}}}}$

= $\displaystyle\lim_{{{x}\to\infty}}\frac{{{\left({{x}}^{{3}}\right)}'}}{{{\left({{e}}^{{\frac{{x}}{{2}}}}\right)}'}}$

= $\displaystyle\lim_{{{x}\to\infty}}\frac{{{3}{{x}}^{{2}}}}{{{\left({{e}}^{{\frac{{x}}{{2}}}}\right)}{\left(\frac{{1}}{{2}}\right)}}}$

again $\displaystyle\frac{{{3}{{x}}^{{2}}}}{{{\left({{e}}^{{\frac{{x}}{{2}}}}\right)}{\left(\frac{{1}}{{2}}\right)}}}$ is of the form $\displaystyle\frac{\infty}{\infty}$ so , applying the L'Hopital Rule we get

= $\displaystyle\lim_{{{x}\to\infty}}\frac{{{3}{{x}}^{{2}}}}{{{\left({{e}}^{{\frac{{x}}{{2}}}}\right)}{\left(\frac{{1}}{{2}}\right)}}}$

= $\displaystyle\lim_{{{x}\to\infty}}\frac{{{\left({3}{{x}}^{{2}}\right)}'}}{{{\left({\left({{e}}^{{\frac{{x}}{{2}}}}\right)}{\left(\frac{{1}}{{2}}\right)}\right)}'}}$

= $\displaystyle\lim_{{{x}\to\infty}}\frac{{{\left({6}{x}\right)}}}{{{\left({\left({{e}}^{{\frac{{x}}{{2}}}}\right)}{\left(\frac{{1}}{{2}}\right)}{\left(\frac{{1}}{{2}}\right)}\right)}}}$

= $\displaystyle\lim_{{{x}\to\infty}}\frac{{{\left({6}{x}\right)}}}{{{\left({\left({{e}}^{{\frac{{x}}{{2}}}}\right)}{\left(\frac{{1}}{{4}}\right)}\right)}}}$

again $\displaystyle\frac{{{\left({6}{x}\right)}}}{{{\left({\left({{e}}^{{\frac{{x}}{{2}}}}\right)}{\left(\frac{{1}}{{4}}\right)}\right)}}}$ is of the form $\displaystyle\frac{\infty}{\infty}$ so , applying the L'Hopital Rule we get

= $\displaystyle\lim_{{{x}\to\infty}}\frac{{{\left({6}{x}\right)}}}{{{\left({\left({{e}}^{{\frac{{x}}{{2}}}}\right)}{\left(\frac{{1}}{{4}}\right)}\right)}}}$

= $\displaystyle\lim_{{{x}\to\infty}}\frac{{{\left({6}{x}\right)}'}}{{{\left({\left({{e}}^{{\frac{{x}}{{2}}}}\right)}{\left(\frac{{1}}{{4}}\right)}\right)}'}}$

= $\displaystyle\lim_{{{x}\to\infty}}\frac{{{\left({6}\right)}}}{{{\left({\left({{e}}^{{\frac{{x}}{{2}}}}\right)}{\left(\frac{{1}}{{4}}\right)}{\left(\frac{{1}}{{2}}\right)}\right)}}}$

= $\displaystyle\lim_{{{x}\to\infty}}\frac{{{\left({6}\right)}}}{{{\left({\left({{e}}^{{\frac{{x}}{{2}}}}\right)}{\left(\frac{{1}}{{8}}\right)}\right)}}}$

= $\displaystyle\lim_{{{x}\to\infty}}\frac{{{\left({6}\cdot{8}\right)}}}{{{\left({\left({{e}}^{{\frac{{x}}{{2}}}}\right)}\right)}}}$

= $\displaystyle\lim_{{{x}\to\infty}}\frac{{{\left({48}\right)}}}{{{\left({\left({{e}}^{{\frac{{x}}{{2}}}}\right)}\right)}}}$

upon plugging the value of $\displaystyle{x}=\infty$

we get

= $\displaystyle\lim_{{{x}\to\infty}}\frac{{{\left({48}\right)}}}{{{\left({\left({{e}}^{{\frac{{\infty}}{{2}}}}\right)}\right)}}}$

= $\displaystyle\lim_{{{x}\to\infty}}\frac{{{\left({48}\right)}}}{{\infty}}$

Notes

Tuesday, 14 February 2017

$\displaystyle\lim_{{{x}\to\infty}}\frac{{{x}}^{{3}}}{{{e}}^{{\frac{{x}}{{2}}}}}$ Evaluate the limit, using L’Hôpital’s Rule if necessary.

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How are race, gender, and class addressed in Oliver Optic's Rich and Humble?

Tuesday, 14 February 2017

Evaluate the limit, using L’Hôpital’s Rule if necessary.

No comments:

Post a Comment

How are race, gender, and class addressed in Oliver Optic&#39;s Rich and Humble?

$\displaystyle\lim_{{{x}\to\infty}}\frac{{{x}}^{{3}}}{{{e}}^{{\frac{{x}}{{2}}}}}$ Evaluate the limit, using L’Hôpital’s Rule if necessary.

How are race, gender, and class addressed in Oliver Optic's Rich and Humble?