Or a much more general question, does the square root of exponent 3 leveling to exponent 2?

If \$t o 7^+\$, the worth of \$t - 7\$ will be always positive, because if \$t\$ is pull close \$7\$ indigenous the ideal side, \$t\$ will constantly be a small bigger 보다 \$7\$, and also thus \$t - 7\$ will constantly be a small bigger 보다 \$0\$ (i.e., positive).

You are watching: Sqrt(7/3)

So simple the limit:

\$lim limits_t o 7^+ dfracsqrt(t-7)^3t-7\$

\$lim limits_t o 7^+ dfracsqrt(t-7)^2sqrt(t-7)t-7\$

\$lim limits_t o 7^+ dfracsqrtt-7t-7\$

Since \$t-7\$ will constantly be positive, \$|t-7|=t-7\$, and also they cancel the end from the numerator and the denominator, leaving:

\$lim limits_t o 7^+ sqrtt-7= 0\$

re-superstructure
cite
follow
answered Sep 4 "15 at 16:35

Ud779Ud779
\$endgroup\$
include a comment |

Thanks for contributing solution to chrischona2015.orgematics Stack Exchange!

Please be certain to answer the question. Provide details and share her research!

But avoid

Asking for help, clarification, or responding to various other answers.Making statements based upon opinion; earlier them up with recommendations or personal experience.

Use chrischona2015.orgJax to layout equations. chrischona2015.orgJax reference.

See more: If You Construct A Line Perpendicular To Line A, Constructing A Perpendicular At A Point On A Line

Draft saved

authorize up making use of Google
submit

Post as a guest

name
email Required, yet never shown

Post together a guest

surname
email

Required, but never shown

Not the answer you're looking for? Browse various other questions tagged algebra-precalculus indices or ask your very own question.

Featured ~ above Meta
associated
3
fix \$sqrt2x-5 - sqrtx-1 = 1\$
2
solving \$fracsqrt108x^10sqrt2x\$
2
follow to Stewart Calculus at an early stage Transcendentals 5th Edition on page 140, in instance 5, exactly how does he simplify this problem?
4
simplify Square source Expression \$sqrt125 - sqrt5\$
0
how to leveling \$sqrt3/2\$?
1
simplify \$sqrt3+sqrt8\$
4
simplify \$sqrtdfracsqrt<3>64 + sqrt<4>256sqrt64+sqrt256\$ right into \$dfracsqrt33\$
1
simplify \$sqrtfracsqrt<4>x^3-8sqrt<4>x-2+2sqrt<4>xleft(fracsqrt<4>x^3+8sqrt<4>x+2-sqrtx ight)\$
warm Network inquiries an ext hot inquiries

question feed