7. Computing Limits

d. Limits at Infinity

3. When Limit Laws Don't Apply

Limits without Laws - Limit Tricks

d. Replace Variable by its Reciprocal

It is frequently useful to change the variable from \(x\) to \(t=\dfrac{1}{x}\) and take the limit as \(t\to0^+\). Notice that this is a limit from the right because as \(x\) goes to positive infinity, \(t=\dfrac{1}{x}\) is positive.

Compute \(\displaystyle \lim_{x\to\infty}x^2\left[1-\cos\left(\dfrac{1}{x}\right)\right]\).

As \(x\) gets large, \(\dfrac{1}{x}\) approaches \(0\), \(\cos\left(\dfrac{1}{x}\right)\) approaches \(1\) and so \(\left[1-\cos\left(\dfrac{1}{x}\right)\right]\) approaches \(0\). So the limit has the indeterminate form \(0\cdot\infty\). We change variables to \(t=\dfrac{1}{x}\) (which changes the indeterminate form to \(\dfrac{0}{0}\)): \[ \lim_{x\to\infty}x^2\left[1-\cos\left(\dfrac{1}{x}\right)\right] =\lim_{t\to 0^+} \dfrac{1-\cos(t)}{t^2} \] We now multiply and divide by the conjugate and use the standard limit \(\displaystyle \lim_{t\to 0} \dfrac{\sin x}{x}=1\): \[\begin{aligned} \lim_{x\to\infty}&x^2\left[1-\cos\left(\dfrac{1}{x}\right)\right] =\lim_{t\to 0^+}\dfrac{1-\cos(t)}{t^2} \dfrac{1+\cos(t)}{1+\cos(t)} \\ &=\lim_{t\to 0^+}\dfrac{1-\cos^2(t)}{t^2(1+\cos(t))} =\lim_{t\to 0^+}\dfrac{\sin^2(t)}{t^2(1+\cos(t))} \\ &=\left(\lim_{t\to 0^+}\dfrac{\sin(t)}{t}\right)^2\lim_{t\to 0^+}\dfrac{1}{1+\cos(t)} =(1)^2\dfrac{1}{1+1}=\dfrac{1}{2} \end{aligned}\]