Evaluate the following limit : lim(x→0) (x(2^x - 1))/(1 - cos x)

Question

Evaluate the following limit : \(\lim\limits_{\text x \to0}\cfrac{\text x(2^{\text x}-1)}{1-cos\,\text x} \)

lim(x→0) (x(2^x - 1))/(1 - cos x)

Answer 1

As we need to find  \(\lim\limits_{\text x \to0}\cfrac{\text x(2^{\text x}-1)}{1-cos\,\text x} \)

lim(x→0) (x(2^x - 1))/(1 - cos x)

We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)

Let Z = \(\lim\limits_{\text x \to0}\cfrac{\text x(2^{\text x}-1)}{1-cos\,\text x} \)

\(=\cfrac00\)(indeterminate form)

∴ We need to take steps to remove this form so that we can get a finite value.

TIP: Most of the problems of logarithmic and exponential limits are solved using the formula

   \(\lim\limits_{\text x \to0}\cfrac{a^{\text x-1}}{\text x}\) = log a and \(\lim\limits_{\text x \to0}\cfrac{log(1+\text x)}{\text x}=1\)

It also involves a trigonometric term, so there is a possibility of application of Sandwich theorem-

\(\lim\limits_{\text x \to0} \cfrac{sin\,\text x}{\text x}=1\)

To get the desired form to apply the formula we need to divide numerator and denominator by x².

Use the formula:    \(\lim\limits_{\text x \to0}\cfrac{a^{\text x-1}}{\text x}\) = log a and \(\lim\limits_{\text x \to0} \cfrac{sin\,\text x}{\text x}=1\)