Determine the limx→∞(ex+x)1x. Use L'Hospital's Rule where appropriate. Use some Elementary method if posible. If L'Hospitals Rule doesn't apply. Explain why.
If we let y=(ex+x)1x, then
lny=(1x)ln(ex+x)
So,
limx→∞lny=limx→∞ln(ex+x)x
By applying L'Hospital's Rule...
limx→∞ln(ex+x)x=limx→∞ex+1ex+x1=limx→∞ex+1ex+x
If we evaluate the limit, we will still get indeterminate form, so we next apply L'Hospital's Rule once more. Thus,
limx→∞ex+1ex+x=limx→∞exex+1
Again, by applying L'Hospital's Rule...
limx→∞exex+1=limx→∞exex=1
Hence,
limx→∞lny=limx→∞ln(ex+x)x=1
Therefore, we have
limx→∞(ex+x)1x=limx→∞elny=e1=e
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