What is #lim_(xrarroo) (e^(2x)sin(1/x))/x^2 #?

Answer 1

#lim_(x->oo) (e^(2x)sin(1/x))/x^2 = oo#

Let #y= (e^(2x)sin(1/x))/x^2#
#lny=ln((e^(2x)sin(1/x))/x^2)#
#lny=lne^(2x)+ln(sin(1/x))-lnx^2#
#lny=2xlne+ln(sin(1/x))-2lnx#
#lny=2x+ln(sin(1/x))-2lnx#
#lim_(x->oo) [lny=2x+ln(sin(1/x))-2lnx]#
#lim_(x->oo)lny = lim_(x->oo)[2x+ln(sin(1/x))-2lnx]#
#lim_(x->oo)lny=oo#
#e^lny=e^oo#
#y=oo#
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Answer 2

#lim_(xrarroo)(e^(2x)sin(1/x))/x^2 = oo#. Please see the explanation section below.

#lim_(xrarroo)(e^(2x)sin(1/x))/x^2#
Note that: #(e^(2x)sin(1/x))/x^2 = e^(2x)/x^3 * sin(1/x)/(1/x)#
Now, as #xrarroo#, the first ratio increases without bound, while the second goes to #1#.
#lim_(xrarroo)(e^(2x)sin(1/x))/x^2 = lim_(xrarroo)e^(2x)/x^3 * lim_(xrarroo)sin(1/x)/(1/x)#
# = oo#

Additional Clarification

This is the logic that produced the previously mentioned solution.

#lim_(xrarroo)(e^(2x)sin(1/x))/x^2# has initial form #(oo*0)/oo#.

Since this is an indeterminate form, l'Hospital's Rule is not applicable.

We could rewrite it as #(e^(2x))/(x^2/sin(1/x))# to get the form #oo/oo# to which we could apply l'Hospital. However, I don't particularly want to take the derivative of that denominator.
Recall that #lim_(thetararr0)sintheta/theta = 1#.
So that #lim_(xrarroo)sin(1/x)/(1/x) = 1#.

This is what drives the rewriting that was done earlier.

#(e^(2x)sin(1/x))/x^2 = e^(2x)/x^3 * sin(1/x)/(1/x)#.
As #x# increases without bound, #e^x# goes to infinity much faster that #x^3# (faster than any power of #x#). So, #e^(2x) = (e^x)^2# blows up even faster.

If you don't have access to this information, apply L'Hospital's rule to determine

#lim_(xrarroo)e^(2x)/x^3 = lim_(xrarroo)(2e^(2x))/(3x^2)#
# = lim_(xrarroo)(8e^(2x))/(6) = oo#
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Answer 3

The limit of (e^(2x)sin(1/x))/x^2 as x approaches infinity is 0.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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