Bypassing the indeterminate form $0\cdot \infty$

Question

Trying to calculate out the limit \begin{array}{rcl} \lim_{x \to +\infty } (e^{x^{2}\sin \frac{1}{x}}-e^{x})&& \\ \end{array}

I come up with the indeterminate form $0\cdot \infty$ as follows: $${e}^{{x}^{2} \sin\frac{1}{x}}-{e}^{x}={e}^{x}\left({e}^{{x}^{2\sin\frac{1}{x}-x}}-1\right)$$ and

\begin{array}{rcl} \lim_{x \to +\infty} (x^2\sin \frac{1}{x}-x) & = & \lim_{x \to \infty}[ x(x\sin \frac{1}{x}-1)] \\ & = & \lim_{x \to \infty} [x (\frac{\sin\frac{1}{x}}{\frac{1}{x}}-1)] \\ & = &( +\infty) \cdot 0 \end{array} And now I am stuck.

Answer 1

Substituting $t=1/x$ gives $$\lim_{x\to\infty} x\left [x^2\sin{1\over x}-x\right ]=\lim_{t\to 0^+}{\sin t-t\over t^3}=-{1\over 6}$$ Let $$g(x):=x^2\sin{1\over x}-x$$ We have $\lim_{x\to \infty}g(x)=0$ and $$ e^x[e^{g(x)}-1]={e^x \over x}\,{xg(x)}\,{e^{g(x)}-1\over g(x)}$$ The last ratio tends to $1,$ while $xg(x)$ tends to $-1/6.$ The ratio $e^x/x$ tends to $\infty.$ Hence the limit in OP is equal $-\infty.$

Answer 2

You can also use L'Hopital's rule to save on the arithmetic. We can change variables as follows:

$$\lim_{x \to \infty} x \left[\frac{\sin(1/x)}{1/x} - 1 \right] = \lim_{x \to \infty}\frac{\frac{\sin(1/x)}{1/x} - 1}{1/x}= \lim_{t \to 0^+}\frac{\frac{\sin(t)}{t} - 1}{t}$$

This is evidently suitable for an application of L'Hopital's rule, so: $$\lim_{t \to 0^+}\frac{\frac{\sin(t)}{t} - 1}{t} = \lim_{t \to 0^+} \frac{d}{dt} \frac{\sin t}{t} = \lim_{t \to 0^+} \frac{t \cos(t) - \sin(t)}{t^2} = \lim_{t \to 0^+} \frac{\cos(t)}{t} - \lim_{t \to 0^+} \frac{\sin(t)}{t^2} = -\infty$$

Answer 3

$\sin x$ for small x is $x - x^3/6 + O(x^5)$.

For large x, $\sin {1/x} = 1/x - 1/{6x^3} + O(1/x^5)$, and $x^2 \sin {1/x} = x - 1/{6x} + O(1/x^3)$.

e raised to this power is $e^x \cdot \exp(-1/6x + O(1/x^3))$.

For small x, $\exp(x) = 1 + x + O(x^2)$.

$\exp(-1/6x + O(1/x^3)) = 1 - 1/{6x} + O(1/x^2)$.

Now we multiply by $e^x$ then subtract $e^x$, and get

$e^x \cdot(-1/6x + O(1/x^2)$.

This is essentially $-e^x / {6x}$ which goes towards minus infinity (calculator on my phone agrees for x=10, result about -365, x=100, result about -4.48 x 10^40).