Limit of the form (0 power 0 or infinity power 0) - Practice Questions & MCQ

Limit of the form 0⁰ or ∞⁰  (0 power 0 or infinity power 0)

In such types of questions, generally, we equate the limit with y, then take the logarithm of both sides. After this, we solve the limits, mostly by using the L’Hospital rule. 

Let’s go through some of the illustrations to understand how such questions are solved

Illustrations 1

The value of $L=\underset{x\to 0^{+}}{lim}(1/x{)}^{\sin x}$ is :
We have an indeterminate form of the type ${\mathrm{\infty }}^0$
Let, $\mathrm{y}=(1/\mathrm{x}{)}^{\sin \mathrm{x}}$
Taking $\log$ on both sides, we get :

$\begin{array}{rl}& \ln y=\sin x\cdot \ln \left(\frac{1}{x}\right)\\ \Rightarrow & \underset{x\to 0^{+}}{lim}\ln y=\underset{x\to 0^{+}}{lim}\sin x\cdot \ln \left(\frac{1}{x}\right)[0\cdot \mathrm{\infty }\text{ form }]\end{array}$
Transform it to $\frac{\mathrm{\infty }}{\mathrm{\infty }}$ form to apply the LH rule.

$\Rightarrow \underset{x\to 0^{+}}{lim}\ln y=\underset{x\to 0^{+}}{lim}\frac{-\ln x}{1/\sin x}$
Apply LH rule

$\begin{array}{rl}& \Rightarrow \underset{x\to 0^{+}}{lim}\ln y=\underset{x\to 0^{+}}{lim}\frac{-1/x}{-(\cos x)/{\sin }^{2}x}\\ & \Rightarrow =\underset{x\to 0^{+}}{lim}\frac{{\sin }^{2}x}{x\cos x}=\underset{x\to 0^{+}}{lim}\frac{\sin x}{x}\cdot \frac{\sin x}{x}\cdot \frac{x}{\cos x}\\ & \Rightarrow \underset{x\to 0^{+}}{lim}y=\underset{x\to 0}{lim}\frac{x}{\cos x}=0\\ & \Rightarrow \underset{x\to 0^{+}}{lim}y=e^0=1\end{array}$