Procedure with limits problem

Question

I've been trying to figure out some limits for about a week and I'm generally having a hard time figuring them out. I don't really know what to look for when encountering such a problem besides making it look like something we know the answer to. I have tried to solve these two limits this past week: \begin{equation} f(x)=\log \sinh(x^2)-x^2 \mbox{ for } x \rightarrow \infty \end{equation} and \begin{equation} f(x)=\frac{\tanh(x)-1}{\mathrm{e}^{-2x}} \mbox{ for } x \rightarrow \infty \end{equation}

If anyone could help me out with any of these or just give me a hint, then it would be greatly appreciated! We were told that we aren't allowed to differentiate or use L'Hopital's rule

Answer 1

Let me help you with the second one, I am not sure of the first one yet.

First, let's rewrite your limit so it looks better:

$$\lim_{x\to\infty}(\tanh(x)-1)e^{2x}$$

Recall that $\tanh(x)-1=\dfrac{e^x-e^{-x}}{e^x+e^{-x}}-1$

We can rewrite using a common denominator

$$=\dfrac{e^x-e^{-x}-[e^x+e^{-x}]}{e^x+e^{-x}}=\dfrac{-2e^{-x}}{e^x+e^{-x}}$$

Now our limit becomes easier to evaluate.

$$\lim_{x\to\infty}\left(\dfrac{-2e^{-x}}{e^x+e^{-x}}\right)e^{2x}$$

Simplifying the top, we get that this limit equals:

$$\lim_{x\to\infty}\left(\dfrac{-2e^x}{e^x+e^{-x}}\right)$$

I trust you can take it from here, it all goes well the answer is $-2$

Answer 2

Here's a hint.

$$\log \sinh x^2 = \log\left(\frac{e^{x^2}+e^{-x^2}}{2}\right)$$ Now $\log$ increases very slowly, and $e^{-x^2}$ goes to $0$ very quickly, so this must be approximately $\log \left(\frac{e^{x^2}}{2}\right)$. Now can you guess the answer, and can you prove it?

Also for part 2. Start by writing down the definition of $\tanh$

Answer 3

For the first note that

$$\sinh x^2=\frac{e^{x^2}-e^{-x^2}}{2}\sim \frac12 e^{x^2}$$

thus

$$\log \sinh(x^2)-x^2 \sim x^2+\log \frac12-x^2\to -\log2$$

For the second

$$\tanh x^2=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}$$

$$\frac{\tanh(x)-1}{e^{-2x}}=e^{2x}\frac{e^{x}-e^{-x}-e^{x}-e^{-x}}{e^{x}+e^{-x}}=e^{2x}\frac{-2e^{-x}}{e^{x}+e^{-x}}=\frac{-2e^{x}}{e^{x}+e^{-x}}\to -2$$

Answer 4

You can use some algebraic manipulation once you rewrite these into exponentials:

$$\sinh(x^2) = \dfrac{e^{x^2}-e^{-x^2}}{2}$$

Then you can factor out an $e^{-x^2}$ from the top:

$$\sinh(x^2) = \dfrac{e^{-x^2}(e^{2(x^2)}-1)}{2}$$ $$\lim_{x\to\infty}\log\left(\dfrac{e^{-x^2}(e^{2(x^2)}-1)}{2}\right)-x^2$$ $$\lim_{x\to\infty}\log\left(\dfrac{e^{2(x^2)}-1}{2e^{x^2}}\right)-x^2$$

Then rewrite $x^2$ as $\log\left(e^{x^2}\right)$:

$$\lim_{x\to\infty}\log\left(\dfrac{e^{2(x^2)}-1}{2e^{x^2}}\right)-\log\left(e^{x^2}\right)$$

Use the log property $\log(a) - \log(b) = \log\left(\dfrac{a}{b}\right)$:

$$\lim_{x\to\infty}\log\left(\dfrac{e^{2(x^2)}-1}{2e^{x^2}e^{x^2}}\right)$$ $$\lim_{x\to\infty}\log\left(\dfrac{e^{2(x^2)}-1}{2e^{x^2}e^{x^2}}\right)$$ $$\lim_{x\to\infty}\log\left(\dfrac{e^{2(x^2)}-1}{2e^{2({x^2})}}\right)$$

From here, you can see as $x$ approaches $\infty$, the $-1$ becomes negligible and you're left with $\log\left(\dfrac{1}{2}\right)$. You can factor out $e^{2(x^2)}$ to see this clearer:

$\require{cancel}$ $$\lim_{x\to\infty}\log\;\left(\dfrac{\cancel{e^{2(x^2)}}\left(1-\cancelto{0}{\dfrac{1}{e^{2(x^2)}}}\right)}{2\cancel{e^{2({x^2})}}}\right)$$

Answer 5

Both the limits can be handled easily if you know the definition of hyperbolic functions in terms of exponential function namely $$\sinh x=\frac{e^x-e^{-x}} {2},\,\tanh x=\frac{e^x-e^{-x}} {e^x+e^{-x}}\tag{1}$$ For the the first limit you can first replace $x^2$ by just $x$ and then note that $$\log\sinh x-x=\log(2e^{-x} \sinh x)-\log 2 =\log(1-e^{-2x})-\log 2$$ Now as $x\to\infty$ the expression $1-e^{-2x}\to 1-0=1$ and hence the first log term above tends to $0$ and therefore the desired limit is $-\log 2$.

For the second limit note that the expression equals $$-\frac{2e^{2x}}{e^{2x}+1}=-\frac{2}{1+e^{-2x}}\to - 2$$ as $x\to \infty$.