Calculus Help

Pre-College Topics High School Life
1

<p>I know that these are probably easy limits, but I just forget everything with limits and we have a test on l'hopital's + series when we get back. I don't think these are l'hopital's problems because they are early in the end of the book review, but I'm not sure. I need to know how to do them algebraically so any help would be great.</p>

<p>lim t(1-cost)
t->0 t-sint
(i couldn't do the line for division, but i think it's clear)</p>

<p>lim ((e^x)+x)^(1/x)
x->0 </p>

<p>lim 3x+1<strong>-</strong>1
x->0 x _____sinx
(just imagine lines underneath the 3x + 1 and the -1 and ignore the underscores... those are only there because CC cut out all the space I had in there before)</p>

<p>Thanks</p>

2

<p>the first one can be solved using l'hopital's, as both top and bottom approach 0.</p>

<p>assuming i read your notation correctly:
lim t->0 (t(1-cost)) / (t - sint)
= lim t->0 (t - tcost)) / (t - sint)
using l'hopital's:
= lim t->0 (1 + tsint - cost) / (1 - cost)
l'hopital's again:
= lim t->0 (sint + tcost + sint) / (sint)
= lim t->0 (2sint + tcost) / (sint)
and again:
= lim t->0 (2cost + cost - tsint) / (cost)
= lim t->0 (3cost - tsint) / (cost)
= 3</p>

<p>ok, number two.</p>

<p>lim x->0 ( e^x + x )^( 1/x )
= e ^ ( limx->0 ln(( e^x + x )^( 1/x )) )
= e ^ ( limx->0 (1/x) (ln( e^x + x )) )
= e ^ ( limx->0 (ln( e^x + x ))) / (x) )
now we have a limit on the inside which heads to 0/0, so we can do l'hopital's
= e ^ ( limx->0 ((e^x + 1) / (e^x + x)) / (1) )
= e ^ ( limx->0 ((e^x + 1) / (e^x + x)) )
= e ^ 2</p>

3

<p>number three</p>

<p>limx->0 [(3x + 1)/(x) - 1/(sinx)]
= limx->0 [((3x + 1)(sinx) - x) / (xsinx) ]
l'hop:
= limx->0 [( 3sinx + (3x + 1)(cosx) - 1) / (sinx + xcosx)]
again:
= limx->0 [( 3cosx + 3cosx - (3x+1)(sinx)) / (cosx + cosx - xsinx)]
= limx->0 [( 6cosx - (3x+1)(sinx)) / (2cosx - xsinx)]
is it over? yes
= 6 / 2
= 3</p>

4

<p>Yeah I concur with all three.</p>

5

<p>Is this how you got to 3,140 posts?</p>

6

<p>You fhore.</p>

7

<p>don't even...</p>

8

<p>Thanks a lot Salt Flakes for solving and the oneo for concurring hahaha</p>