Find the value of $c$ such that $\lim_{x\to\infty} \frac{1+ce^x}{\sqrt{1+cx^2}} = 4$

Question

Find the value of $c$ such that $$\lim_{x\to\infty} \frac{1+ce^x}{\sqrt{1+cx^2}} = 4$$ I don't think if there exists such $c$, but don't know how to prove.

Answer 1

Yes, there exists no such value of $\color{red}{c}$ However, You can prove these types results by proving that it's independent of whatever parameters are mentioned there in the equation(In your question if you think there exists no such $\color{red}{c}$ then you simply need to prove it's independent of $\color{red}{c}$)

Also, We need to know only when $x\to\infty$

Here, I guess $c ≠ 0$ As this adds no meaning for applying limit and for $x\to\infty$ negative values of $\color{red}c$ will compel denominator to give complex roots but numerator will have negative sign only {$\frac{-ve}{a+ib} \in \mathbb C∉ \mathbb R ≠ 4 $ } can't be $4$(or Real)

Conclusion: We'll worry for only +ve values of $\color{red}c$ for which Numerator and denominator will tend to $\infty$ and we can use L'Hôpital's rule

$$\begin{align*}\lim_{x\to\infty}\frac{1+ce^x}{\sqrt{1+cx^2}} & = \lim_{x\to\infty}\frac{ce^x}{\frac{2xc}{2\sqrt{1+cx^2}}}\\ \text{L'Hôpital's rule as type }\left(\frac{\infty}{\infty}\right) & = \lim_{x\to\infty}\frac{e^x\sqrt{1+cx^2}}{x}\\ & = \lim_{x\to\infty}e^x\sqrt{\frac{1}{x^2}+\color{red}{c}}\\ & = \text{As, } x\to\infty \text{ limit of the function } \to \infty \text { independent of } \color{red}{c}. \end{align*}$$

Answer 2

You're correct that such a $c$ doesn't exist. Here's how to prove it:

Notice that this limit only makes sense if $c$ non-negative; if $c<0$, then the expression $\sqrt{1+cx^2}$ would not be defined for every sufficiently large $x$, rendering the limit meaningless.

With this established, note that we can write

\begin{align} \frac{1+ce^x}{\sqrt{1+cx^2}} &= \frac{1}{\sqrt{1+cx^2}}+\frac{ce^x}{\sqrt{1+cx^2}}\\ &= \frac{1}{\sqrt{1+cx^2}}+\frac{c}{\frac{1}{e^x}\sqrt{1+cx^2}}\\ &= \frac{1}{\sqrt{1+cx^2}}+\frac{c}{\sqrt{\frac{1}{e^{2x}}\left(1+cx^2\right)}}\\ &= \frac{1}{\sqrt{1+cx^2}}+\frac{c}{\sqrt{\frac{1+cx^2}{e^{2x}}}}\\ \end{align}

so if $c=0$, we have

$$\frac{1+ce^x}{\sqrt{1+cx^2}}=\frac{1}{\sqrt{1+0}}+\frac{0}{\sqrt{\frac{1+0}{e^{2x}}}}=1$$

and thus the limit is $1$. If $c>0$, $\frac{1}{\sqrt{1+cx^2}}\to 0$ and $$\frac{c}{\sqrt{\frac{1+cx^2}{e^{2x}}}}\to\infty\text{ because }\sqrt{\frac{1+cx^2}{e^{2x}}}\to 0$$ so

$$\lim_{x\to\infty}\frac{1+ce^x}{\sqrt{1+cx^2}}=\infty$$

Thus, the limit can never be $4$.

Answer 3

First of all, it is clear that if $c<0$, then the expression in the limit will be negative whenever $|c|e^x>1$ which happens whenever $x>-\ln |c|$.

It is also clear that if $c=0$, the limit is $1$.

For $c>0$, you can use the following estimate to see where the limit will go:

$$\begin{align} \frac{1+ce^x}{\sqrt{1+cx^2}} &= \frac{\frac{1+ce^x}{x}}{\frac{\sqrt{1+cx^2}}{x}}\\ &=\frac{\frac{1+ce^x}{x}}{\sqrt{\frac{1}{x^2}+c}}\\ &\geq \frac{1}{\sqrt{c+1}}\cdot\frac{ce^x}{x}\\ &\geq \frac{c}{\sqrt{c+1}}\cdot \frac{e^x}{x} \end{align}$$

After you do this estimate, you can use the fact that $$\lim_{x\to\infty}\frac{e^x}{x}=\infty$$ which can be proven in a number of ways.

Answer 4

$e^x=\Omega(x^3)$ so that your expression is $\Omega\left(\sqrt c \,x\right)$, which diverges. ($c=0$ is not a solution.)