New Notes Blog: `lim_(x->oo)sinx/(x-pi)` Evaluate the limit, using L’Hôpital’s Rule if necessary.

Wednesday, 15 January 2014

$\displaystyle\lim_{{{x}\to\infty}}\frac{{\sin{{x}}}}{{{x}-\pi}}$ Evaluate the limit, using L’Hôpital’s Rule if necessary.

Given to solve,

$\displaystyle\lim_{{{x}\to\infty}}\frac{{\sin{{x}}}}{{{x}-\pi}}$

This can be solved by applying the squeeze theorem and is as follows

as we know the limits or boundaries of $\displaystyle{\sin{{\left({x}\right)}}}$ is

$\displaystyle-{1}\le{\sin{{\left({x}\right)}}}\le{1}$

Dividing the above expression with x-pi we get

$\displaystyle-\frac{{1}}{{{x}-\pi}}\le\frac{{\sin{{\left({x}\right)}}}}{{{x}-\pi}}\le\frac{{1}}{{{x}-\pi}}$

now, let us apply the limits that $\displaystyle{x}\to\infty$ we get

$\displaystyle\lim_{{{x}\to\infty}}{\left(-\frac{{1}}{{{x}-\pi}}\right)}\le\lim_{{{x}\to\infty}}\frac{{\sin{{\left({x}\right)}}}}{{{x}-\pi}}\le\lim_{{{x}\to\infty}}\frac{{1}}{{{x}-\pi}}$

but

$\displaystyle\lim_{{{x}\to\infty}}{\left(-\frac{{1}}{{{x}-\pi}}\right)}=-\frac{{1}}{{\infty-\pi}}={0}$

$\displaystyle\lim_{{{x}\to\infty}}{\left(\frac{{1}}{{{x}-\pi}}\right)}=\frac{{1}}{{\infty-\pi}}={0}$

so,

$\displaystyle{0}\le\lim_{{{x}\to\infty}}\frac{{\sin{{\left({x}\right)}}}}{{{x}-\pi}}\le{0}$

so ,

$\displaystyle\lim_{{{x}\to\infty}}\frac{{\sin{{\left({x}\right)}}}}{{{x}-\pi}}={0}$

Given to solve,

$\displaystyle\lim_{{{x}\to\infty}}\frac{{\sin{{x}}}}{{{x}-\pi}}$

This can be solved by applying the squeeze theorem and is as follows

as we know the limits or boundaries of $\displaystyle{\sin{{\left({x}\right)}}}$ is

$\displaystyle-{1}\le{\sin{{\left({x}\right)}}}\le{1}$

Dividing the above expression with x-pi we get

$\displaystyle-\frac{{1}}{{{x}-\pi}}\le\frac{{\sin{{\left({x}\right)}}}}{{{x}-\pi}}\le\frac{{1}}{{{x}-\pi}}$

now, let us apply the limits that $\displaystyle{x}\to\infty$ we get

$\displaystyle\lim_{{{x}\to\infty}}{\left(-\frac{{1}}{{{x}-\pi}}\right)}\le\lim_{{{x}\to\infty}}\frac{{\sin{{\left({x}\right)}}}}{{{x}-\pi}}\le\lim_{{{x}\to\infty}}\frac{{1}}{{{x}-\pi}}$

but

$\displaystyle\lim_{{{x}\to\infty}}{\left(-\frac{{1}}{{{x}-\pi}}\right)}=-\frac{{1}}{{\infty-\pi}}={0}$

$\displaystyle\lim_{{{x}\to\infty}}{\left(\frac{{1}}{{{x}-\pi}}\right)}=\frac{{1}}{{\infty-\pi}}={0}$

so,

$\displaystyle{0}\le\lim_{{{x}\to\infty}}\frac{{\sin{{\left({x}\right)}}}}{{{x}-\pi}}\le{0}$

so ,

$\displaystyle\lim_{{{x}\to\infty}}\frac{{\sin{{\left({x}\right)}}}}{{{x}-\pi}}={0}$

New Notes Blog

Wednesday, 15 January 2014

$\displaystyle\lim_{{{x}\to\infty}}\frac{{\sin{{x}}}}{{{x}-\pi}}$ Evaluate the limit, using L’Hôpital’s Rule if necessary.

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In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?

Wednesday, 15 January 2014

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

No comments:

Post a Comment

In &quot;By the Waters of Babylon,&quot; under the leadership of John, what do you think the Hill People will do with their society?

$\displaystyle\lim_{{{x}\to\infty}}\frac{{\sin{{x}}}}{{{x}-\pi}}$ Evaluate the limit, using L’Hôpital’s Rule if necessary.

In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?