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Limit of $lim_x to inftyfracC_3a^xD_1+(ka)^xD_2$

1

$begingroup$

I want to calculate the following limit

$$lim_x to inftyfracC_3a^xD_1+(ka)^xD_2$$

knowing that $-1<a<1$, $-1<k<1$ and $1<A<2$.

I proceed in the following way.

Since for $x to infty$ we have $left[frac00right]$, I use the de L'Hopital's rule.

$$lim_x to inftyfracC_3a^xD_1+(ka)^xD_2=left[frac00right]=lim_x to inftyfracC_1a^x+C_2(ka)^xa^xlog(a)D_1+(ka)^xlog(ka)D_2\frac^A-1sgn(C_1a^x+C_2(ka)^x-C_3)(C_1a^xlog(a)+C_2(ka)^xlog(ka))a^xlog(a)D_1+(ka)^xlog(ka)D_2$$

Now, I take the expression $a^x$ in front of both nominator and denominator and reduce it obtaining

$$lim_x to inftyfracAlog(a)D_1+(k)^xlog(ka)D_2\frac^A-1sgn(C_1a^x+C_2(ka)^x-C_3)(C_1log(a)+C_2(k)^xlog(ka))log(a)D_1+(k)^xlog(ka)D_2$$

and finally I obtain $fracAlog(a)D_1$ as the result.

Is this reasoning correct?

My main question is:

At some point, I calculate the derivative of $a^x$ which is equal to $a^xlog(a)$ and I am sure that it is correct in the case of $0<a<1$, but what if $-1<a<0$? Can I write that the derivative of $a^x$ is $a^xlog(a)$ when $-1<a<0$? If not, how should I calculate this limit in the case of $-1<a<0$?

real-analysis calculus limits derivatives

share|cite|improve this question

edited Mar 25 at 15:58

Gatey

asked Mar 25 at 9:40

GateyGatey

163

$endgroup$

add a comment | 

1

$begingroup$

I want to calculate the following limit

$$lim_x to inftyfracC_3a^xD_1+(ka)^xD_2$$

knowing that $-1<a<1$, $-1<k<1$ and $1<A<2$.

I proceed in the following way.

Since for $x to infty$ we have $left[frac00right]$, I use the de L'Hopital's rule.

$$lim_x to inftyfracC_3a^xD_1+(ka)^xD_2=left[frac00right]=lim_x to inftyfracC_1a^x+C_2(ka)^xa^xlog(a)D_1+(ka)^xlog(ka)D_2\frac^A-1sgn(C_1a^x+C_2(ka)^x-C_3)(C_1a^xlog(a)+C_2(ka)^xlog(ka))a^xlog(a)D_1+(ka)^xlog(ka)D_2$$

Now, I take the expression $a^x$ in front of both nominator and denominator and reduce it obtaining

$$lim_x to inftyfracAlog(a)D_1+(k)^xlog(ka)D_2\frac^A-1sgn(C_1a^x+C_2(ka)^x-C_3)(C_1log(a)+C_2(k)^xlog(ka))log(a)D_1+(k)^xlog(ka)D_2$$

and finally I obtain $fracAlog(a)D_1$ as the result.

Is this reasoning correct?

My main question is:

At some point, I calculate the derivative of $a^x$ which is equal to $a^xlog(a)$ and I am sure that it is correct in the case of $0<a<1$, but what if $-1<a<0$? Can I write that the derivative of $a^x$ is $a^xlog(a)$ when $-1<a<0$? If not, how should I calculate this limit in the case of $-1<a<0$?

real-analysis calculus limits derivatives

share|cite|improve this question

edited Mar 25 at 15:58

Gatey

asked Mar 25 at 9:40

GateyGatey

163

$endgroup$

add a comment | 

$begingroup$

I want to calculate the following limit

$$lim_x to inftyfracC_3a^xD_1+(ka)^xD_2$$

knowing that $-1<a<1$, $-1<k<1$ and $1<A<2$.

I proceed in the following way.

Since for $x to infty$ we have $left[frac00right]$, I use the de L'Hopital's rule.

$$lim_x to inftyfracC_3a^xD_1+(ka)^xD_2=left[frac00right]=lim_x to inftyfracC_1a^x+C_2(ka)^xa^xlog(a)D_1+(ka)^xlog(ka)D_2\frac^A-1sgn(C_1a^x+C_2(ka)^x-C_3)(C_1a^xlog(a)+C_2(ka)^xlog(ka))a^xlog(a)D_1+(ka)^xlog(ka)D_2$$

Now, I take the expression $a^x$ in front of both nominator and denominator and reduce it obtaining

$$lim_x to inftyfracAlog(a)D_1+(k)^xlog(ka)D_2\frac^A-1sgn(C_1a^x+C_2(ka)^x-C_3)(C_1log(a)+C_2(k)^xlog(ka))log(a)D_1+(k)^xlog(ka)D_2$$

and finally I obtain $fracAlog(a)D_1$ as the result.

Is this reasoning correct?

My main question is:

At some point, I calculate the derivative of $a^x$ which is equal to $a^xlog(a)$ and I am sure that it is correct in the case of $0<a<1$, but what if $-1<a<0$? Can I write that the derivative of $a^x$ is $a^xlog(a)$ when $-1<a<0$? If not, how should I calculate this limit in the case of $-1<a<0$?

real-analysis calculus limits derivatives

share|cite|improve this question

edited Mar 25 at 15:58

Gatey

asked Mar 25 at 9:40

GateyGatey

163

$endgroup$

share|cite|improve this question

edited Mar 25 at 15:58

Gatey

asked Mar 25 at 9:40

GateyGatey

163

share|cite|improve this question

edited Mar 25 at 15:58

Gatey

asked Mar 25 at 9:40

GateyGatey

163

asked Mar 25 at 9:40

GateyGatey

163

asked Mar 25 at 9:40

GateyGatey

163