Limits

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au
and

Department of Mathematics
University of Wisconsin, Madison
Madison, WI 53706 USA
ram@math.wisc.edu

Last updates: 27 October 2009

Limits

Write limx→2fx=10 if fx gets closer and closer to 10 as x gets closer and closer to 2.

Example. Evaluate limx→23x2+8x2-x.
When x=1.5, 3x2+8x2-x=19.66⋯.
When x=1.9, 3x2+8x2-x=11.011⋯.
When x=1.99, 3x2+8x2-x=10.091⋯.
When x=1.999, 3x2+8x2-x=10.00901⋯.
When x=1.9999, 3x2+8x2-x=10.0009001⋯.

Usually determining the limit is straightforward.

Example. limx→16x2-4x+3=5.

But sometimes …

Example. limx→11+x-1x=?00.
Hoewver 00 MAKES NO SENSE.

Example. limx→05xx=?00.
However here we have limx→05xx=limx→05=5.

Example. limx→017x2x=?00.
However here we have limx→017x2x=limx→0172=172.

Example. limx→11+x-1x=?00.
However here we have limx→01+x-1x=limx→01+x-1x·1+x+11+x+1=limx→01+x-1x1+x+1=limx→0xx1+x+1=limx→011+x+1=11+0+1=12. So whenever a limit looks like it is coming out to 00 it needs to be looked at in a different way to see what it is really getting closer and closer to.

Example. Evaluate limx→7x2-49x-7.
Here we have limx→7x2-49x-7=limx→7x-7x+7x-7=limx→7x+7=7+7=14.

Example. Evaluate limx→5x5-3125x-5.
Here we have limx→5x5-3125x-5=limx→5x5-55x-5=limx→5x-5x4+5x3+52x2+53x+54x-5=limx→5x4+5x3+52x2+53x+54=54+54+54+54+54=55=3125.

Example. Evaluate limx→5x5/2-a5/2x-a.
Here we have limx→5x5/2-a5/2x-a=limx→5x5/2-a5/2x-a·x5/2+a5/2x5/2+a5/2=limx→5x5-a5x-a·1x5/2+a5/2=limx→5x-ax4+ax3+a2x2+a3x+a4x-a·1x5/2+a5/2=limx→5x4+ax3+a2x2+a3x+a4x5/2+a5/2=a4+a4+a4+a4+a4a5/2+a5/2=5a42a5/2=52a3/2.

Particularly useful limits

Example. Evaluate limx→0sinxx.
Here we have limx→0sinxx=limx→0x-x33!+x55!-x77!+x99!-⋯x=limx→01-x23!+x45!-x67!+x89!-⋯=1-0+0-0+0-⋯=1

Example. Evaluate limx→0cosx-1x.
Here we have limx→0cosx-1x=limx→01-x22!+x44!-x66!+x88!-⋯-1x=limx→0-x22!+x44!-x66!+x88!-⋯x=limx→0-x2!+x34!-x56!+x78!-⋯=-0+0-0+0-⋯=0.

Example. Evaluate limx→0ex-1x.
Here we have limx→0ex-1x=limx→01+x+x22!+x33!+x44!+⋯-1x=limx→0x+x22!+x33!+x44!+⋯x=limx→01+x2!+x23!+x34!+⋯=1+0+0+0+⋯=1.

Example. Evaluate limx→0ln1+xx.
Let y=ln1+x. Then ey=1+x⇒x=ey-1,andy→0 as x→0. So limx→0ln1+xx=limy→0yey-1=limy→01ey-1y=11=1.

Example. Evaluate limx→01+x1/x.
We have limx→01+x1/x=limx→0eln1+x1/x=limx→0e1xln1+x=limx→0eln1+xx=e1=e.

Note. n→∞ means as n gets larger and larger.

Example. Evaluate limn→∞1+1nn.
Let x=1n. Then x→0 as n→∞. So limn→∞1+1nn=limx→01+x1/x=e..

Example. Evaluate limx→πsinxx-π.
Let y=x-π. Then y→0 as x→π. So limx→πsinxx-π=limy→0siny+πy=limy→0sinycosπ+cosysinπy=limy→0siny-1+cosy0y=limy→0-sinyy=1

Example. Evaluate limx→∞x2 7x+113x2+10.
We have limx→∞x2-7x+113x2+10=limx→∞1-7x+11x23+10x2=1-0+03+0=13.

Example. Evaluate limx→0sin3xsin5x.
We have limx→0sin3xsin5x=limx→0sin3x3x·3x·5xsin5x·15x=limx→0sin3x3x·1sin5x5x·3x5x=1·11·35=35.

Example. Evaluate limx→11-xarccosx2.
Let y=arccosx. Then y→0 as x→1 and x=cosy. So limx→11-xarccosx2=limy→01-cosyy2=limy→01-cosyy2·1+cosy1+cosy=limy→01-cos2yy2·11+cosy=limy→0sinyy·sinyy·11+cosy=1·1·12=12.

Example. Evaluate limΔx→0fx+Δx-fxΔx when fx=sin2x.
We have limΔx→0fx+Δx-fxΔx=limΔx→0sin2x+Δx-sin2xΔx=limΔx→0sin2x+2Δx-sin2xΔx=limΔx→0sin2xcos2Δx+cos2xsin2Δx-sin2xΔx=limΔx→0sin2x·cos2Δx-1Δx+cos2x·sin2ΔxΔx=limΔx→0sin2x·cos2Δx-12Δx·2+cos2x·sin2Δx2Δx·2=sin2x·0·2+cos2x·1·2=2cos2x.

Example. Evaluate limΔx→0fx+Δx-fxΔx when fx=cosx2.
We have limΔx→0fx+Δx-fxΔx=limΔx→0cosx+Δx2-cosx2Δx=limΔx→0cosx2+2xΔx+Δx2-cosx2Δx=limΔx→0cosx2cos2xΔx+Δx2-sinx2sin2xΔx+Δx2-cosx2Δx=limΔx→0cosx2·cos2xΔx+Δx2-1Δx-sinx2·sin2xΔx+Δx2Δx=limΔx→0cosx2·cos2xΔx+Δx2-12xΔx+Δx2·2xΔx+Δx2Δx-sinx2·sin2xΔx+Δx22xΔx+Δx2·2xΔx+Δx2Δx=limΔx→0cosx2·cos2xΔx+Δx2-12xΔx+Δx2·2x+Δx-sinx2·sin2xΔx+Δx22xΔx+Δx2·2x+Δx=cosx2·0·2x-sinx2·1·2x=-2xsinx2, (since 2xΔx+Δx2→0 as Δx→0).

Example. Evaluate limΔx→0fx+Δx-fxΔx when fx=xx.
We have limΔx→0fx+Δx-fxΔx=limΔx→0x+Δxx+Δx+xxΔx=limΔx→0elnx+Δxx+Δx+elnxxΔx=limΔx→0ex+Δxlnx+Δx+exlnxΔx=limΔx→0exlnx·ex+Δxlnx+Δx-xlnx+1Δx=limΔx→0exlnx·ex+Δxlnx+Δx-xlnx+1x+Δxlnx+Δx-xlnx·x+Δxlnx+Δx-xlnxΔx=limΔx→0exlnx·ex+Δxlnx+Δx-xlnx+1x+Δxlnx+Δx-xlnxxlnx+Δx-xlnxΔx+lnx+Δ x=limΔx→0exlnx·ex+Δxlnx+Δx-xlnx+1x+Δxlnx+Δx-xlnxxlnx1+Δxx-xlnxΔx+lnx+Δ x=limΔx→0exlnx·ex+Δxlnx+Δx-xlnx+1x+Δxlnx+Δx-xlnxxlnx+ln1+Δxx-xlnxΔx+lnx+Δ x=limΔx→0exlnx·ex+Δxlnx+Δx-xlnx+1x+Δxlnx+Δx-xlnxxln1+ΔxxΔx+lnx+Δ x=limΔx→0exlnx·ex+Δxlnx+Δx-xlnx+1x+Δxlnx+Δx-xlnxln1+ΔxxΔxx+lnx+Δ x=exlnx·1·1+lnx=xx+xxlnx. (since x+Δxlnx+Δx-xlnx→0 and Δx/x→0 as Δx→0).

References [PLACEHOLDER]

[BG] A. Braverman and D. Gaitsgory, Crystals via the affine Grassmanian, Duke Math. J. 107 no. 3, (2001), 561-575; arXiv:math/9909077v2, MR1828302 (2002e:20083)