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Verifying the limit of the expressions
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I have calculated the limit of the following expressions as follows:
$$lim_x rightarrow 0 fracx times frac2x5x + 3 + e^-2x = lim_x rightarrow 0 frac2x5x^2 + 3x + xe^-2x = lim_x rightarrow 0 frac210x + 3 + e^-2x -2xe^-2x (textusing Le Chatelier's principle) = frac23+1$$
$$lim_x rightarrow 0 fracx times frac2x4 + frac1-x1+x = lim_x rightarrow 0 frac2x(1+x)4x(1+x) + x(1-x) = lim_x rightarrow 0 frac2x(1+x)5x-3x^2 = lim_x rightarrow 0 frac4x+25 - 6x (textusing Le Chatelier's principle) = frac25$$
Are my calculations correct?
calculus
asked Jul 18 at 19:58
Eval
17810
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up vote
1
down vote
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I have calculated the limit of the following expressions as follows:
$$lim_x rightarrow 0 fracx times frac2x5x + 3 + e^-2x = lim_x rightarrow 0 frac2x5x^2 + 3x + xe^-2x = lim_x rightarrow 0 frac210x + 3 + e^-2x -2xe^-2x (textusing Le Chatelier's principle) = frac23+1$$
$$lim_x rightarrow 0 fracx times frac2x4 + frac1-x1+x = lim_x rightarrow 0 frac2x(1+x)4x(1+x) + x(1-x) = lim_x rightarrow 0 frac2x(1+x)5x-3x^2 = lim_x rightarrow 0 frac4x+25 - 6x (textusing Le Chatelier's principle) = frac25$$
Are my calculations correct?
calculus
asked Jul 18 at 19:58
Eval
17810
2
Yes, but why didn't you just simplify the numerators?
– Bernard
Jul 18 at 20:01
@Eval Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/…
– gimusi
Aug 8 at 23:05
add a comment |Â
up vote
1
down vote
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up vote
1
down vote
favorite
I have calculated the limit of the following expressions as follows:
$$lim_x rightarrow 0 fracx times frac2x5x + 3 + e^-2x = lim_x rightarrow 0 frac2x5x^2 + 3x + xe^-2x = lim_x rightarrow 0 frac210x + 3 + e^-2x -2xe^-2x (textusing Le Chatelier's principle) = frac23+1$$
$$lim_x rightarrow 0 fracx times frac2x4 + frac1-x1+x = lim_x rightarrow 0 frac2x(1+x)4x(1+x) + x(1-x) = lim_x rightarrow 0 frac2x(1+x)5x-3x^2 = lim_x rightarrow 0 frac4x+25 - 6x (textusing Le Chatelier's principle) = frac25$$
Are my calculations correct?
calculus
asked Jul 18 at 19:58
Eval
17810
I have calculated the limit of the following expressions as follows:
$$lim_x rightarrow 0 fracx times frac2x5x + 3 + e^-2x = lim_x rightarrow 0 frac2x5x^2 + 3x + xe^-2x = lim_x rightarrow 0 frac210x + 3 + e^-2x -2xe^-2x (textusing Le Chatelier's principle) = frac23+1$$
$$lim_x rightarrow 0 fracx times frac2x4 + frac1-x1+x = lim_x rightarrow 0 frac2x(1+x)4x(1+x) + x(1-x) = lim_x rightarrow 0 frac2x(1+x)5x-3x^2 = lim_x rightarrow 0 frac4x+25 - 6x (textusing Le Chatelier's principle) = frac25$$
Are my calculations correct?
calculus
asked Jul 18 at 19:58
Eval
17810
edited Jul 18 at 20:08
asked Jul 18 at 19:58
Eval
17810
asked Jul 18 at 19:58
Eval
17810
asked Jul 18 at 19:58
Eval
17810
17810
2
Yes, but why didn't you just simplify the numerators?
– Bernard
Jul 18 at 20:01
@Eval Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/…
– gimusi
Aug 8 at 23:05
add a comment |Â
2
Yes, but why didn't you just simplify the numerators?
– Bernard
Jul 18 at 20:01
@Eval Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/…
– gimusi
Aug 8 at 23:05
2
2
Yes, but why didn't you just simplify the numerators?
– Bernard
Jul 18 at 20:01
Yes, but why didn't you just simplify the numerators?
– Bernard
Jul 18 at 20:01
@Eval Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/…
– gimusi
Aug 8 at 23:05
@Eval Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/…
– gimusi
Aug 8 at 23:05
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
As suggested by Bernard in the above comment, more simply we have
$$lim_x rightarrow 0 fraccolorredx times frac2colorredx5x + 3 + e^-2x = lim_x rightarrow 0 frac25x + 3 + e^-2x stackreltextby continuity= frac23+1=frac12$$
and similarly for the second we have
$$lim_x rightarrow 0 fraccolorredx times frac2colorredx4 + frac1-x1+x =lim_x rightarrow 0 frac24 + frac1-x1+x stackreltextby continuity= frac24+1=frac25$$
Refer also to: Why are we allowed to cancel fractions in limits?
answered Jul 18 at 20:07
gimusi
65.4k73584
We can cancel out x in the numerator and the denominator. Right?
– Eval
Jul 18 at 20:10
Yes of course, I've highlighted that in red.
– gimusi
Jul 18 at 20:11
Will the solution still be correct if x is a complex number?
– Eval
Jul 18 at 20:13
@Eval Yes in this case the solution should be the same, indeed we can simplify in teh same way and $5zto 0$, $e^-2z=e^-2xe^-2iyto1$, $frac1-z1+zto 1$.
– gimusi
Jul 18 at 20:25
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
As suggested by Bernard in the above comment, more simply we have
$$lim_x rightarrow 0 fraccolorredx times frac2colorredx5x + 3 + e^-2x = lim_x rightarrow 0 frac25x + 3 + e^-2x stackreltextby continuity= frac23+1=frac12$$
and similarly for the second we have
$$lim_x rightarrow 0 fraccolorredx times frac2colorredx4 + frac1-x1+x =lim_x rightarrow 0 frac24 + frac1-x1+x stackreltextby continuity= frac24+1=frac25$$
Refer also to: Why are we allowed to cancel fractions in limits?
answered Jul 18 at 20:07
gimusi
65.4k73584
We can cancel out x in the numerator and the denominator. Right?
– Eval
Jul 18 at 20:10
Yes of course, I've highlighted that in red.
– gimusi
Jul 18 at 20:11
Will the solution still be correct if x is a complex number?
– Eval
Jul 18 at 20:13
@Eval Yes in this case the solution should be the same, indeed we can simplify in teh same way and $5zto 0$, $e^-2z=e^-2xe^-2iyto1$, $frac1-z1+zto 1$.
– gimusi
Jul 18 at 20:25
add a comment |Â
up vote
1
down vote
accepted
As suggested by Bernard in the above comment, more simply we have
$$lim_x rightarrow 0 fraccolorredx times frac2colorredx5x + 3 + e^-2x = lim_x rightarrow 0 frac25x + 3 + e^-2x stackreltextby continuity= frac23+1=frac12$$
and similarly for the second we have
$$lim_x rightarrow 0 fraccolorredx times frac2colorredx4 + frac1-x1+x =lim_x rightarrow 0 frac24 + frac1-x1+x stackreltextby continuity= frac24+1=frac25$$
Refer also to: Why are we allowed to cancel fractions in limits?
answered Jul 18 at 20:07
gimusi
65.4k73584
We can cancel out x in the numerator and the denominator. Right?
– Eval
Jul 18 at 20:10
Yes of course, I've highlighted that in red.
– gimusi
Jul 18 at 20:11
Will the solution still be correct if x is a complex number?
– Eval
Jul 18 at 20:13
@Eval Yes in this case the solution should be the same, indeed we can simplify in teh same way and $5zto 0$, $e^-2z=e^-2xe^-2iyto1$, $frac1-z1+zto 1$.
– gimusi
Jul 18 at 20:25
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
As suggested by Bernard in the above comment, more simply we have
$$lim_x rightarrow 0 fraccolorredx times frac2colorredx5x + 3 + e^-2x = lim_x rightarrow 0 frac25x + 3 + e^-2x stackreltextby continuity= frac23+1=frac12$$
and similarly for the second we have
$$lim_x rightarrow 0 fraccolorredx times frac2colorredx4 + frac1-x1+x =lim_x rightarrow 0 frac24 + frac1-x1+x stackreltextby continuity= frac24+1=frac25$$
Refer also to: Why are we allowed to cancel fractions in limits?
answered Jul 18 at 20:07
gimusi
65.4k73584
As suggested by Bernard in the above comment, more simply we have
$$lim_x rightarrow 0 fraccolorredx times frac2colorredx5x + 3 + e^-2x = lim_x rightarrow 0 frac25x + 3 + e^-2x stackreltextby continuity= frac23+1=frac12$$
and similarly for the second we have
$$lim_x rightarrow 0 fraccolorredx times frac2colorredx4 + frac1-x1+x =lim_x rightarrow 0 frac24 + frac1-x1+x stackreltextby continuity= frac24+1=frac25$$
Refer also to: Why are we allowed to cancel fractions in limits?
answered Jul 18 at 20:07
gimusi
65.4k73584
edited Jul 18 at 20:11
answered Jul 18 at 20:07
gimusi
65.4k73584
answered Jul 18 at 20:07
gimusi
65.4k73584
answered Jul 18 at 20:07
gimusi
65.4k73584
65.4k73584
We can cancel out x in the numerator and the denominator. Right?
– Eval
Jul 18 at 20:10
Yes of course, I've highlighted that in red.
– gimusi
Jul 18 at 20:11
Will the solution still be correct if x is a complex number?
– Eval
Jul 18 at 20:13
@Eval Yes in this case the solution should be the same, indeed we can simplify in teh same way and $5zto 0$, $e^-2z=e^-2xe^-2iyto1$, $frac1-z1+zto 1$.
– gimusi
Jul 18 at 20:25
add a comment |Â
We can cancel out x in the numerator and the denominator. Right?
– Eval
Jul 18 at 20:10
Yes of course, I've highlighted that in red.
– gimusi
Jul 18 at 20:11
Will the solution still be correct if x is a complex number?
– Eval
Jul 18 at 20:13
@Eval Yes in this case the solution should be the same, indeed we can simplify in teh same way and $5zto 0$, $e^-2z=e^-2xe^-2iyto1$, $frac1-z1+zto 1$.
– gimusi
Jul 18 at 20:25
We can cancel out x in the numerator and the denominator. Right?
– Eval
Jul 18 at 20:10
We can cancel out x in the numerator and the denominator. Right?
– Eval
Jul 18 at 20:10
Yes of course, I've highlighted that in red.
– gimusi
Jul 18 at 20:11
Yes of course, I've highlighted that in red.
– gimusi
Jul 18 at 20:11
Will the solution still be correct if x is a complex number?
– Eval
Jul 18 at 20:13
Will the solution still be correct if x is a complex number?
– Eval
Jul 18 at 20:13
@Eval Yes in this case the solution should be the same, indeed we can simplify in teh same way and $5zto 0$, $e^-2z=e^-2xe^-2iyto1$, $frac1-z1+zto 1$.
– gimusi
Jul 18 at 20:25
@Eval Yes in this case the solution should be the same, indeed we can simplify in teh same way and $5zto 0$, $e^-2z=e^-2xe^-2iyto1$, $frac1-z1+zto 1$.
– gimusi
Jul 18 at 20:25
add a comment |Â
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2
Yes, but why didn't you just simplify the numerators?
– Bernard
Jul 18 at 20:01
@Eval Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/…
– gimusi
Aug 8 at 23:05