Solving $\lim_{x\to0^+}e^{1/x}\bigl(1-\sec(x)\bigr)$ algebraically

Question

I want to solve the following limit using algebraic techniques learnt in calculus class (including l'Hospital rule): $$\lim_{x\to0^+}e^{1/x}\bigl(1-\sec(x)\bigr)$$

I have tried simplifying the expression and putting over the same denomniator.

$$\lim_{x\to0^+}\frac{e^{1/x}(\cos(x)-1)}{\cos(x)}$$

Then I just need to solve this expression: $$\lim_{x\to0^+}e^{1/x}(\cos(x)-1)$$

I have tried expressing it as a fraction to apply l'Hospital's rule but it only seems to get worst. $$\lim_{x\to0^+}\frac{\cos(x)-1}{e^{-1/x}}=\lim_{x\to0^+}\frac{-x^2\sin(x)}{e^{-1/x}}$$

Answer 1

I have solved it, using $u\to\infty,\;\; u=\frac1x$ (I don't think the transformation is necessary). Then apply sandwich theorem with $-2e^u \leq e^{u}(\cos(1/u)-1) \leq0$.

Update: This solution does not quite work, see comments.

Answer 2

Use $$\sec(x)=1+\frac{x^2}{2}+O\left(x^4\right)$$ which makes the expression to be $$-\frac{1}{2} e^{\frac{1}{x}} x^2$$ Let $x=\frac 1y$ $$-\frac{e^y}{2 y^2}$$ which is $\frac \infty \infty$

Use twice L'Hospital rule to see what happens when $y\to \infty$.

Answer 3

What you need to do, starting from $\lim_{x \to 0^+} e^{\frac 1x} (\cos(x)-1)$ is look for an expression $g(x)$ such that $\frac{\cos(x)-1}{g(x)}$ has a finite non-zero limit as $x \to 0^+$, and then multiply and divide by $g(x)$. Isolate the term $\frac{\cos(x)-1}{g(x)}$ whose limit you know is non-zero and finite as $x \to 0^+$, and then deal with $g(x)e^{\frac 1x}$ separately. The reason this helps is that doing L'Hospital blindly leads to the recurrence of the term $e^{\frac 1x}$ in every such term which means you don't really end up getting a simpler limit when you differentiate.

For example, tempted by the fact that the cosine will admit a Taylor expansion at the point $0$, we try a function of the form $g(x) = x^n$ where $n$ is a natural number (frankly, this works in a huge number of situations). Then, we consider $\lim_{x \to 0+}\frac{\cos(x)-1}{x^n}$ and attempt to L'Hospitalize this, getting in the process :$$ \frac{\cos(x)-1}{x^n} \to \frac{-\sin(x)}{nx^{n-1}} \to \frac{-\cos(x)}{n(n-1)x^{n-2}} $$ Now the numerator converges to $-1$ as $x \to 0^+$, so if the denominator converges to a non-zero quantity as $x \to 0^+$ then the L'Hospital rule will tell us that the first limit is some non-zero finite number. That only occurs when $n-2=0$ or $n=2$. That is, we must have $g(x) = x^2$ and therefore have the decomposition $$ e^{\frac 1x}(\cos(x)-1) = e^{\frac 1x}x^2 \times \frac{\cos(x)-1}{x^2}. $$ Now, the latter's limit can be found by L'Hospital and is a negative quantity, as we discovered earlier.

How about the first limit? Here, the point is to not L'Hospitalize blindly : try it by setting up a $\frac 00$ situation and noticing that there is some potential to fail. Instead, it is wiser to change the variable and simplify the exponential, which you do by setting $y = \frac 1x$ so that you're dealing with $$ \lim_{x \to 0^+} e^{\frac 1x}x^2 = \lim_{y \to \infty} \frac{e^y}{y^2} $$ whereupon, L'Hospitalizing twice leads to the answer $+\infty$. Thus, the answer is $+\infty$ times something negative, which is $-\infty$. The answer to the very first limit in this answer is also $-\infty$, keeping in mind the work done in the original post.

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Someone (outside of this site) asked me : what if we instead looked for a function $g(x)$ such that $\frac{e^\frac 1x}{g(x)}$ had a limit as $x \to 0^+$, and then found the limit $g(x)(\cos(x)-1)$ instead? Why are we "preferring" the function $\cos(x)-1$ over the function $e^{\frac 1x}$ while multiplying and dividing by $g(x)$.

Someone taking that approach may feel stuck though. To explain this, observe that if one tries to take e.g. $g(x)= x^n$ for any integer $n$ and tries to compute $$ \lim_{x \to 0^+} \frac{e^{\frac 1x}}{x^n} $$ Then the first thing to do is to substitute $y = \frac 1x$ to get $\lim_{y \to +\infty} e^{y}y^{n}$. Turns out that regardless of the value of $n$, this limit is infinite (as one can see by performing L'Hospital as many times as necessary). Therefore, students may feel stuck when they encounter this, but the key point is that one doesn't need to associate $g(x)$ to $e^{\frac 1x}$ but can associate it to $\cos(x)-1$ instead and that happens to work out.