# Indeterminate Forms

## Solved Problems

### Example 5.

Find the limit $\lim\limits_{x \to \pi } {\frac{{\cos \frac{x}{2}}}{{\pi - x}}}.$

Solution.

Change the variable: $$x - \pi = t$$ or $$x = t + \pi.$$ Then $$t \to 0$$ as $$x \to \pi.$$ We have

$L = \lim\limits_{x \to \pi } \frac{{\cos \frac{x}{2}}}{{\pi - x}} = \left[ {\frac{0}{0}} \right] = \lim\limits_{t \to 0} \frac{{\cos \frac{{t + \pi }}{2}}}{{ - t}}.$

Convert the last expression using the reduction formula $$\cos\left( {z + \frac{\pi }{2}} \right) = - \sin z.$$ As a result we find the limit of the function:

$L = \lim\limits_{t \to 0} \frac{{\cos \frac{{t + \pi }}{2}}}{{ - t}} = \lim\limits_{t \to 0} \frac{{ - \cos \frac{t}{2}}}{{ - t}} = \lim\limits_{\frac{t}{2} \to 0} \frac{{\sin \frac{t}{2}}}{{\frac{{2t}}{2}}} = \frac{1}{2}\lim\limits_{\frac{t}{2} \to 0} \frac{{\sin \frac{t}{2}}}{{\frac{t}{2}}} = \frac{1}{2} \cdot 1 = \frac{1}{2}.$

### Example 6.

Calculate $\lim\limits_{x \to \infty } \left( {\sqrt {{x^2} + 1} - \sqrt {{x^2} - 1} } \right).$

Solution.

If $$x \to \infty,$$ then

$\lim\limits_{x \to \infty } \sqrt {{x^2} + 1} = \infty \;\;\; \text{and}\;\;\lim\limits_{x \to \infty } \sqrt {{x^2} - 1} = \infty .$

Thus, we deal here with an indeterminate form of type $$\infty - \infty.$$ Multiply this expression (both the numerator and the denominator) by the corresponding conjugate expression.

$L = \lim\limits_{x \to \infty } \left( {\sqrt {{x^2} + 1} - \sqrt {{x^2} - 1} } \right) = \lim\limits_{x \to \infty } \frac{{{{\left( {\sqrt {{x^2} + 1} } \right)}^2} - {{\left( {\sqrt {{x^2} - 1} } \right)}^2}}}{{\left( {\sqrt {{x^2} + 1} + \sqrt {{x^2} - 1} } \right)}} = \lim\limits_{x \to \infty } \frac{{{x^2} + 1 - \left( {{x^2} - 1} \right)}}{{\left( {\sqrt {{x^2} + 1} + \sqrt {{x^2} - 1} } \right)}} = \lim\limits_{x \to \infty } \frac{{\cancel{x^2} + 1 - \cancel{x^2} + 1}}{{\left( {\sqrt {{x^2} + 1} + \sqrt {{x^2} - 1} } \right)}} = \lim\limits_{x \to \infty } \frac{2}{{\left( {\sqrt {{x^2} + 1} + \sqrt {{x^2} - 1} } \right)}}.$

By using the product and the sum rules for limits, we obtain

$L = \lim\limits_{x \to \infty } \frac{2}{{\left( {\sqrt {{x^2} + 1} + \sqrt {{x^2} - 1} } \right)}} = \frac{{\lim\limits_{x \to \infty } 2}}{{\lim\limits_{x \to \infty } \sqrt {{x^2} + 1} + \lim\limits_{x \to \infty } \sqrt {{x^2} - 1} }} \sim \frac{2}{{\infty + \infty }} \sim \frac{2}{\infty } = 0.$

### Example 7.

Find the limit $\lim\limits_{x \to 4} {\frac{{\sqrt {1 + 6x} - 5}}{{\sqrt x - 2}}}.$

Solution.

To calculate this limit we rationalize the numerator and denominator multiplying them by the corresponding conjugate expressions:

$\lim\limits_{x \to 4} \frac{{\sqrt {1 + 6x} - 5}}{{\sqrt x - 2}} = \left[ {\frac{0}{0}} \right] = \lim\limits_{x \to 4} \frac{{\left( {1 + 6x - 25} \right)\left( {\sqrt x + 2} \right)}}{{\left( {\sqrt {1 + 6x} + 5} \right)\left( {x - 4} \right)}} = \lim\limits_{x \to 4} \frac{{6 \cancel{\left( {x - 4} \right)} \left( {\sqrt x + 2} \right)}}{{\left( {\sqrt {1 + 6x} + 5} \right) \cancel{\left( {x - 4} \right)}}} = 6\lim\limits_{x \to 4} \frac{{\sqrt x + 2}}{{\sqrt {1 + 6x} + 5}} = 6 \cdot \frac{{\sqrt 4 + 2}}{{\sqrt {25} + 5}} = 6 \cdot \frac{4}{{10}} = \frac{{12}}{5}.$

### Example 8.

Find the limit $\lim\limits_{x \to \infty } {\frac{{{{\left( {2x + 3} \right)}^{10}}{{\left( {3x - 2} \right)}^{20}}}}{{{{\left( {x + 5} \right)}^{30}}}}}.$

Solution.

We divide both the numerator and denominator by $${x^{30}}$$ (the highest power of the fraction):

$\lim\limits_{x \to \infty } \frac{{{{\left( {2x + 3} \right)}^{10}}{{\left( {3x - 2} \right)}^{20}}}}{{{{\left( {x + 5} \right)}^{30}}}} = \left[ {\frac{\infty }{\infty }} \right] = \lim\limits_{x \to \infty } \frac{{\frac{{{{\left( {2x + 3} \right)}^{10}}{{\left( {3x - 2} \right)}^{20}}}}{{{x^{30}}}}}}{{\frac{{{{\left( {x + 5} \right)}^{30}}}}{{{x^{30}}}}}} = \lim\limits_{x \to \infty } \frac{{\frac{{{{\left( {2x + 3} \right)}^{10}}}}{{{x^{10}}}} \cdot \frac{{{{\left( {3x - 2} \right)}^{20}}}}{{{x^{20}}}}}}{{\frac{{{{\left( {x + 5} \right)}^{30}}}}{{{x^{30}}}}}} = \lim\limits_{x \to \infty } \frac{{{{\left( {\frac{{2x + 3}}{x}} \right)}^{10}} \cdot {{\left( {\frac{{3x - 2}}{x}} \right)}^{20}}}}{{{{\left( {\frac{{x + 5}}{x}} \right)}^{30}}}} = \lim\limits_{x \to \infty } \frac{{{{\left( {2 + \frac{3}{x}} \right)}^{10}} \cdot {{\left( {3 - \frac{2}{x}} \right)}^{20}}}}{{{{\left( {1 + \frac{5}{x}} \right)}^{30}}}} = \frac{{{2^{10}} \cdot {3^{20}}}}{{{1^{30}}}} = {2^{10}} \cdot {3^{10}} \cdot {3^{10}} = {18^{10}}.$

### Example 9.

Find the limit $\lim\limits_{x \to e} {\frac{{\ln x - 1}}{{x - e}}}.$

Solution.

Let $$x - e = t.$$ Then $$t \to 0$$ as $$x \to e.$$ Hence,

$\lim\limits_{x \to e} \frac{{\ln x - 1}}{{x - e}} = \left[ {\frac{0}{0}} \right] = \lim\limits_{t \to 0} \frac{{\ln \left( {t + e} \right) - 1}}{t} = \lim\limits_{t \to 0} \frac{{\ln \left( {t + e} \right) - \ln e}}{t} = \lim\limits_{t \to 0} \left( {\frac{1}{t}\ln \frac{{t + e}}{e}} \right) = \lim\limits_{t \to 0} \ln {\left( {\frac{{t + e}}{e}} \right)^{\frac{1}{t}}} = \left[ {{1^\infty }} \right] = \lim\limits_{t \to 0} \ln \left[ {{{\left( {1 + \frac{t}{e}} \right)}^{\frac{e}{t} \cdot \frac{1}{e}}}} \right] = \lim\limits_{t \to 0} \left[ {\frac{1}{e}\ln {{\left( {1 + \frac{t}{e}} \right)}^{\frac{e}{t}}}} \right] = \frac{1}{e}\ln \left[ {\lim\limits_{\frac{t}{e} \to 0} {{\left( {1 + \frac{t}{e}} \right)}^{\frac{e}{t}}}} \right] = \frac{1}{e} \ln e = \frac{1}{e}.$

### Example 10.

Find the limit $\lim\limits_{t \to + \infty } \left( {\sqrt {t + \sqrt {t + 1} } - \sqrt t } \right).$

Solution.

This function is defined only for $$t \ge 0.$$ Multiply and divide it by the conjugate expression $$\left( {\sqrt {t + \sqrt {t + 1} } + \sqrt t } \right).$$ So, we get

$L = \lim\limits_{t \to + \infty } \left( {\sqrt {t + \sqrt {t + 1} } - \sqrt t } \right) = \left[ {\infty - \infty } \right] = \lim\limits_{t \to + \infty } \frac{{{{\left( {\sqrt {t + \sqrt {t + 1} } } \right)}^2} - {{\left( {\sqrt t } \right)}^2}}}{{\sqrt {t + \sqrt {t + 1} } + \sqrt t }} = \lim\limits_{t \to + \infty } \frac{{\cancel{t} + \sqrt {t + 1} - \cancel{t}}}{{\sqrt {t + \sqrt {t + 1} } + \sqrt t }} = \lim\limits_{t \to + \infty } \frac{{\sqrt {t + 1} }}{{\sqrt {t + \sqrt {t + 1} } + \sqrt t }} = \left[ {\frac{\infty }{\infty }} \right].$

Both the numerator and denominator now approach $$\infty$$ as $$t \to \infty.$$ Hence, we divide numerator and denominator by $$t^{1/2},$$ the highest power of $$t$$ in the denominator. Then

$L = \lim\limits_{t \to + \infty } \frac{{\sqrt {t + 1} }}{{\sqrt {t + \sqrt {t + 1} } + \sqrt t }} = \lim\limits_{t \to + \infty } \frac{{\frac{{\sqrt {t + 1} }}{{\sqrt t }}}}{{\frac{{\sqrt {t + \sqrt {t + 1} } + \sqrt t }}{{\sqrt t }}}} = \lim\limits_{t \to + \infty } \frac{{\sqrt {\frac{{t + 1}}{t}} }}{{\sqrt {\frac{{t + \sqrt {t + 1} }}{t}} + 1}} = \lim\limits_{t \to + \infty } \frac{{\sqrt {1 + \frac{1}{t}} }}{{\sqrt {1 + \sqrt {\frac{1}{t} + \frac{1}{{{t^2}}}} } + 1}} = \frac{{\sqrt 1 }}{{\sqrt 1 + 1}} = \frac{1}{2}.$