Indeterminate Forms for Functions

Example 1. Find $\lim_{{{x}\to\infty}}{\left({3}{{x}}^{{2}}-{5}{x}\right)}$.

Since $\lim_{{{x}\to\infty}}{3}{{x}}^{{2}}=\infty$ and $\lim_{{{x}\to\infty}}{5}{x}=\infty$ we have indeterminate form of type $\infty-\infty$.

To handle it, let's perform algebraic manipulations: ${3}{{x}}^{{2}}-{5}{x}={{x}}^{{2}}{\left({3}-\frac{{5}}{{x}}\right)}$.

Now, since $\lim_{{{x}\to\infty}}{{x}}^{{2}}=\infty$ and $\lim_{{{x}\to\infty}}{\left({3}-\frac{{5}}{{x}}\right)}={3}$ then their product is also very large number: $\lim_{{{x}\to\infty}}{\left({3}{{x}}^{{2}}-{5}{x}\right)}=\infty$.

Example 2. Find $\lim_{{{x}\to\infty}}{\left(-{4}{{x}}^{{3}}+{5}{{x}}^{{2}}\right)}$.

Since $\lim_{{{x}\to\infty}}-{4}{{x}}^{{3}}=-\infty$ and $\lim_{{{x}\to\infty}}{5}{{x}}^{{2}}=\infty$ we have indeterminate form of type $\infty-\infty$.

To handle it, let's perform algebraic manipulations: $-{4}{{x}}^{{2}}+{5}{{x}}^{{2}}={{x}}^{{3}}{\left(-{4}+\frac{{5}}{{x}}\right)}$.

Now, since $\lim_{{{x}\to\infty}}{{x}}^{{3}}=\infty$ and $\lim_{{{x}\to\infty}}{\left(-{4}+\frac{{5}}{{x}}\right)}=-{4}$ then $\lim_{{{x}\to\infty}}{\left(-{4}{{x}}^{{3}}+{5}{x}\right)}=-\infty$.

Example 3. Find $\lim_{{{x}\to\infty}}{\left({a}_{{0}}{{x}}^{{k}}+{a}_{{1}}{{x}}^{{{k}-{1}}}+\ldots+{a}_{{{k}-{1}}}{x}+{a}_{{k}}\right)}$, where ${a}_{{0}},{a}_{{1}},\ldots{a}_{{k}}$ are constants.

This is generalization of above two examples. If all coefficients ${a}_{{0}},{a}_{{1}},\ldots,{a}_{{k}}$ have same sign then limit of this sequence is $\infty$ (or $-\infty$). But if coeffcients have different signs then we have indeterminate form of type $\infty-\infty$.

To handle it, let's perform algebraic manipulations: ${a}_{{0}}{{x}}^{{k}}+{a}_{{1}}{{x}}^{{{k}-{1}}}+\ldots+{a}_{{{k}-{1}}}{x}+{a}_{{k}}={{x}}^{{k}}{\left({a}_{{0}}+\frac{{{a}_{{1}}}}{{x}}+\ldots+\frac{{{a}_{{{k}-{1}}}}}{{{{x}}^{{{k}-{1}}}}}+\frac{{{a}_{{k}}}}{{{{x}}^{{k}}}}\right)}$.

Now, since ${{x}}^{{k}}\to\infty$ and ${a}_{{0}}+\frac{{{a}_{{1}}}}{{x}}+\ldots+\frac{{{a}_{{{k}-{1}}}}}{{{{x}}^{{{k}-{1}}}}}+\frac{{{a}_{{k}}}}{{{x}}^{{k}}}\to{a}_{{0}}$ as ${x}\to\infty$, then ${a}_{{0}}{{x}}^{{k}}+{a}_{{1}}{{x}}^{{{k}-{1}}}+\ldots+{a}_{{{k}-{1}}}{x}+{a}_{{k}}\to\infty$ if ${a}_{{0}}>{0}$ and ${x}_{{n}}\to-\infty$ if ${a}_{{0}}<{0}$.

Example 4. Find $\lim_{{{x}\to\infty}}\frac{{{3}{{x}}^{{2}}-{5}{x}}}{{{7}{x}+{3}}}$.

Since $\lim_{{{x}\to\infty}}{\left({3}{{x}}^{{2}}-{5}{x}\right)}\to\infty$ and $\lim_{{{x}\to\infty}}{\left({7}{x}+{3}\right)}\to\infty$ we have indeterminate form of type $\frac{{\infty}}{{\infty}}$.

To handle it, let's perform algebraic manipulations. Factor out ${x}$ raised to the greatest degree in numerator and denominator (in this case ${{x}}^{{2}}$): $\frac{{{3}{{x}}^{{2}}-{5}{x}}}{{{7}{x}+{3}}}=\frac{{{{x}}^{{2}}{\left({3}-\frac{{5}}{{x}}\right)}}}{{{{x}}^{{2}}{\left(\frac{{7}}{{x}}+\frac{{3}}{{{x}}^{{2}}}\right)}}}=\frac{{{3}-\frac{{5}}{{{x}}^{{2}}}}}{{\frac{{7}}{{x}}+\frac{{3}}{{{x}}^{{2}}}}}$.

Now, since ${3}-\frac{{5}}{{x}}\to{3}$ and $\frac{{7}}{{x}}+\frac{{3}}{{{x}}^{{2}}}\to{0}$ as ${x}\to\infty$, then $\frac{{{3}-\frac{{5}}{{{x}}^{{2}}}}}{{\frac{{7}}{{x}}+\frac{{3}}{{{x}}^{{2}}}}}\to\infty$.

Example 5. Let $\lim_{{{x}\to\infty}}\frac{{{6}{{x}}^{{4}}-{3}{{x}}^{{2}}}}{{{8}{{x}}^{{7}}+{3}{x}}}$.

Since $\lim_{{{x}\to\infty}}{\left({6}{{x}}^{{4}}-{3}{{x}}^{{2}}\right)}\to\infty$ and $\lim_{{{x}\to\infty}}{\left({8}{{x}}^{{7}}+{3}{x}\right)}\to\infty$ we have indeterminate form of type $\frac{{\infty}}{{\infty}}$.

To handle it, let's perform algebraic manipulations. Factor out ${x}$ raised to the greatest degree in numerator and denominator (in this case ${{x}}^{{7}}$): $\frac{{{6}{{x}}^{{4}}-{3}{{x}}^{{2}}}}{{{8}{{x}}^{{7}}+{3}{x}}}=\frac{{{{x}}^{{7}}{\left(\frac{{6}}{{{x}}^{{3}}}-\frac{{3}}{{{x}}^{{5}}}\right)}}}{{{{x}}^{{7}}{\left({8}+\frac{{3}}{{{x}}^{{6}}}\right)}}}=\frac{{\frac{{6}}{{{x}}^{{3}}}-\frac{{3}}{{{x}}^{{5}}}}}{{{8}+\frac{{3}}{{{x}}^{{6}}}}}$.

Now, since $\frac{{6}}{{{x}}^{{3}}}-\frac{{3}}{{{x}}^{{5}}}\to{0}$ and ${8}+\frac{{3}}{{{x}}^{{6}}}\to{8}$ as ${x}\to\infty$, then $\lim_{{{x}\to\infty}}\frac{{\frac{{6}}{{{x}}^{{3}}}-\frac{{3}}{{{x}}^{{5}}}}}{{{8}+\frac{{3}}{{{x}}^{{6}}}}}\to{0}$.

Example 6. Find $\lim_{{{x}\to\infty}}\frac{{{3}{{x}}^{{2}}-{5}{x}}}{{{7}{{x}}^{{2}}+{3}}}$.

Since $\lim_{{{x}\to\infty}}{\left({3}{{x}}^{{2}}-{5}{x}\right)}=\infty$ and $\lim_{{{x}\to\infty}}{\left({7}{{x}}^{{2}}+{3}\right)}=\infty$ we have indeterminate form of type $\frac{{\infty}}{{\infty}}$.

To handle it, let's perform algebraic manipulations. Factor out ${x}$ raised to the greatest degree in numerator and denominator (in this case ${{x}}^{{2}}$) $\frac{{{3}{{x}}^{{2}}-{5}{x}}}{{{7}{{x}}^{{2}}+{3}}}=\frac{{{{x}}^{{2}}{\left({3}-\frac{{5}}{{x}}\right)}}}{{{{n}}^{{2}}{\left({7}+\frac{{3}}{{{x}}^{{2}}}\right)}}}=\frac{{{3}-\frac{{5}}{{x}}}}{{{7}+\frac{{3}}{{{x}}^{{2}}}}}$.

Now, since ${3}-\frac{{5}}{{x}}\to{3}$ and ${7}+\frac{{3}}{{{x}}^{{2}}}\to{7}$ as ${x}\to\infty$, then $\lim_{{{x}\to\infty}}\frac{{{3}-\frac{{5}}{{x}}}}{{{7}+\frac{{3}}{{{x}}^{{2}}}}}=\frac{{3}}{{7}}$.

Example 7. Find $\lim_{{{x}\to\infty}}\frac{{{a}_{{0}}{{x}}^{{k}}+{a}_{{1}}{{x}}^{{{k}-{1}}}+\ldots+{a}_{{{k}-{1}}}{x}+{a}_{{k}}}}{{{b}_{{0}}{{x}}^{{m}}+{b}_{{1}}{{x}}^{{{m}-{1}}}+..+{b}_{{{m}-{1}}}{x}+{b}_{{m}}}}$ where ${a}_{{0}},{a}_{{1}},\ldots,{a}_{{k}}$ and ${b}_{{0}},{b}_{{1}},\ldots,{b}_{{m}}$ are constants.

This is generalization of above three examples. We have indeterminate form of type $\frac{{\infty}}{{\infty}}$.

To handle it, let's perform algebraic manipulations. Factor out ${{x}}^{{k}}$ from numerator and ${{x}}^{{m}}$ from denominator: $\frac{{{a}_{{0}}{{x}}^{{k}}+{a}_{{1}}{{x}}^{{{k}-{1}}}+\ldots+{a}_{{{k}-{1}}}{x}+{a}_{{k}}}}{{{b}_{{0}}{{x}}^{{m}}+{b}_{{1}}{{x}}^{{{m}-{1}}}+\ldots+{b}_{{{m}-{1}}}{x}+{b}_{{m}}}}=\frac{{{{x}}^{{k}}{\left({a}_{{0}}+\frac{{{a}_{{1}}}}{{x}}+\ldots+\frac{{{a}_{{k}}}}{{{{x}}^{{k}}}}\right)}}}{{{{x}}^{{m}}{\left({b}_{{0}}+\frac{{{b}_{{1}}}}{{x}}+\ldots+\frac{{{b}_{{m}}}}{{{{x}}^{{m}}}}\right)}}}={{x}}^{{{k}-{m}}}{\left(\frac{{{a}_{{0}}+\frac{{{a}_{{1}}}}{{x}}+\ldots+\frac{{{a}_{{k}}}}{{{{x}}^{{k}}}}}}{{{b}_{{0}}+\frac{{{b}_{{1}}}}{{x}}+\ldots+\frac{{{b}_{{m}}}}{{{{x}}^{{m}}}}}}\right)}$.

Second factor has limit $\frac{{a}_{{0}}}{{b}_{{0}}}$. If ${k}={m}$ then ${{x}}^{{{k}-{m}}}={1}\to{1}$ and required limit approaches $\frac{{{a}_{{0}}}}{{{b}_{{0}}}}$. If ${k}>{m}$ then ${{x}}^{{{k}-{m}}}\to\infty$ and required limit approaches $\infty$ (or $-\infty$, sign depends on sign of $\frac{{{a}_{{0}}}}{{{b}_{{0}}}}$). If ${k}<{m}$ then ${{x}}^{{{k}-{m}}}\to{0}$ and required limit approaches ${0}$.

Example 8. Find $\lim_{{{x}\to\infty}}\sqrt{{{x}}}{\left(\sqrt{{{x}+{1}}}-\sqrt{{{x}}}\right)}$.

We can rewrite it as $\lim_{{{x}\to\infty}}{\left(\sqrt{{{x}}}\sqrt{{{x}+{1}}}-\sqrt{{{x}}}\sqrt{{{x}}}\right)}=\lim_{{{x}\to\infty}}{\left(\sqrt{{{{x}}^{{2}}+{x}}}-{x}\right)}$.

This indeterminate form of type $\infty-\infty$.

When we deal with radicals, we need to multiply both numerator and denominator by conjugate radical. Here we multiply both numerator and denominator by $\sqrt{{{{x}}^{{2}}+{x}}}+{x}$:

$\sqrt{{{{x}}^{{2}}+{x}}}-{x}=\frac{{{\left(\sqrt{{{{x}}^{{2}}+{x}}}-{x}\right)}{\color{red}{{{\left(\sqrt{{{{x}}^{{2}}+{x}}}+{x}\right)}}}}}}{{{\color{red}{{\sqrt{{{{x}}^{{2}}+{x}}}+\sqrt{{{x}}}}}}}}=\frac{{{{\left(\sqrt{{{{x}}^{{2}}+{x}}}\right)}}^{{2}}-{{x}}^{{2}}}}{{\sqrt{{{{x}}^{{2}}+{x}}}+\sqrt{{{x}}}}}=\frac{{{{x}}^{{2}}+{x}-{{x}}^{{2}}}}{{\sqrt{{{{x}}^{{2}}+{x}}}+{x}}}=$

$=\frac{{{x}}}{{\sqrt{{{{x}}^{{2}}+{x}}}+{x}}}$.

Now, factor out ${x}$: $\frac{{x}}{{\sqrt{{{{x}}^{{2}}+{x}}}+{x}}}={\left(\frac{{x}}{{{x}{\left(\sqrt{{{1}+\frac{{1}}{{x}}}}+{1}\right)}}}\right)}=\frac{{1}}{{\sqrt{{{1}+\frac{{1}}{{x}}}}-{1}}}$.

Since ${1}<\sqrt{{{1}+\frac{{1}}{{x}}}}<{1}+\frac{{1}}{{x}}$ for large ${x}$ and ${1}+\frac{{1}}{{x}}\to{1}$ as ${x}\to\infty$ then by Squeeze Theorem $\sqrt{{{1}+\frac{{1}}{{x}}}}\to{1}$.

That's why $\lim_{{{x}\to\infty}}{\left(\sqrt{{{{x}}^{{2}}+{x}}}-{x}\right)}=\lim_{{{x}\to\infty}}{\left(\frac{{1}}{{\sqrt{{{1}+\frac{{1}}{{x}}}}-{1}}}\right)}=\frac{{1}}{{{1}+{1}}}=\frac{{1}}{{2}}$.

Now, we will calculate limit that we will need later.

Example 9. Find $\lim_{{{x}\to{0}}}\frac{{{\sin{{\left({x}\right)}}}}}{{x}}$.

As ${x}\to{0}$ ${\sin{{\left({x}\right)}}}\to{0}$, so we have indeterminate form of type $\frac{{0}}{{0}}$.

Before finding this limit we need to prove that for ${0}<{x}<\frac{\pi}{{2}}$ we have that ${\sin{{\left({x}\right)}}}<{x}<{\tan{{\left({x}\right)}}}$.

For this consider circle with radius ${r}$, tangent ${A}{C}$ to circle and central angle ${x}$.

Clearly area of triangle AOB is less than area of sector AOB, and area of sector AOB is less than area of triangle AOC.

We have that area of sector AOB is $\frac{{{x}}}{{{2}\pi}}\cdot\pi{{r}}^{{2}}=\frac{{1}}{{2}}{{r}}^{{2}}{x}$.

Since area of triangle AOB is $\frac{{1}}{{2}}{{r}}^{{2}}{\sin{{\left({x}\right)}}}$, area of sector AOB is $\frac{{1}}{{2}}{{r}}^{{2}}{x}$ and area of triangle AOC is $\frac{{1}}{{2}}{{r}}^{{2}}{\tan{{\left({x}\right)}}}$ then $\frac{{1}}{{2}}{{r}}^{{2}}{\sin{{\left({x}\right)}}}<\frac{{1}}{{2}}{{r}}^{{2}}{x}<\frac{{1}}{{2}}{{r}}^{{2}}{\tan{{\left({x}\right)}}}$ or ${\sin{{\left({x}\right)}}}<{x}<{\tan{{\left({x}\right)}}}$.

Now, since ${0}<{x}<\frac{\pi}{{2}}$ then we can divide ${\sin{{\left({x}\right)}}}$ by each member of inequality (don't forget to invert inequality signs): $\frac{{{\sin{{\left({x}\right)}}}}}{{{\color{red}{{{\sin{{\left({x}\right)}}}}}}}}>\frac{{{\sin{{\left({x}\right)}}}}}{{{\color{red}{{{x}}}}}}>\frac{{{\sin{{\left({x}\right)}}}}}{{{\color{red}{{{\tan{{\left({x}\right)}}}}}}}}$ or ${1}>\frac{{{\sin{{\left({x}\right)}}}}}{{x}}>{\cos{{\left({x}\right)}}}$.

Thus, we obtained that ${\cos{{\left({x}\right)}}}<\frac{{{\sin{{\left({x}\right)}}}}}{{x}}<{1}$.

As ${x}\to{0}$ ${\cos{{\left({x}\right)}}}\to{1}$ that's why $\lim_{{{x}\to{0}}}{\cos{{\left({x}\right)}}}={1}$, so, by Squeeze Theorem $\lim_{{{x}\to{{0}}^{+}}}\frac{{{\sin{{\left({x}\right)}}}}}{{x}}={1}$. (we wrote one-sided limit because we considered interval ${\left({0},\frac{\pi}{{2}}\right)}$). Since $\frac{{{\sin{{\left({x}\right)}}}}}{{x}}$ is even function then also $\lim_{{{x}\to{{0}}^{{-}}}}\frac{{{\sin{{\left({x}\right)}}}}}{{x}}={1}$. Since one-sided limits are equal then $\lim_{{{x}\to{0}}}\frac{{{\sin{{\left({x}\right)}}}}}{{x}}={1}$.

From this limit we can find other limits.

Example 10. Find $\lim_{{{x}\to{0}}}\frac{{{1}-{\cos{{\left({x}\right)}}}}}{{{x}}^{{2}}}$.

Here we have indeterminate form of type $\frac{{0}}{{0}}$.

Since ${1}-{\cos{{\left({x}\right)}}}={2}{{\sin}}^{{2}}{\left(\frac{{x}}{{2}}\right)}$ then $\frac{{{1}-{\cos{{\left({x}\right)}}}}}{{{x}}^{{2}}}=\frac{{{2}{{\sin}}^{{2}}{\left(\frac{{x}}{{2}}\right)}}}{{{x}}^{{2}}}=\frac{{1}}{{2}}\frac{{{{\sin}}^{{2}}{\left(\frac{{x}}{{2}}\right)}}}{{\frac{{{{x}}^{{2}}}}{{4}}}}=\frac{{1}}{{2}}{{\left(\frac{{{\sin{{\left(\frac{{x}}{{2}}\right)}}}}}{{\frac{{x}}{{2}}}}\right)}}^{{2}}$.

Since ${x}\to{0}$ then $\frac{{x}}{{2}}\to{0}$ and so $\lim_{{{x}\to{0}}}\frac{{{\sin{{\left(\frac{{x}}{{2}}\right)}}}}}{{\frac{{x}}{{2}}}}={1}$.

Thus, $\lim_{{{x}\to{0}}}\frac{{{1}-{\cos{{\left({x}\right)}}}}}{{{x}}^{{2}}}=\lim_{{{x}\to{0}}}\frac{{1}}{{2}}{{\left(\frac{{{\sin{{\left(\frac{{x}}{{2}}\right)}}}}}{{\frac{{x}}{{2}}}}\right)}}^{{2}}=\frac{{1}}{{2}}{{\left(\lim_{{{x}\to{0}}}\frac{{{\sin{{\left(\frac{{x}}{{2}}\right)}}}}}{{\frac{{x}}{{2}}}}\right)}}^{{2}}=\frac{{1}}{{2}}\cdot{{1}}^{{2}}=\frac{{1}}{{2}}$.

When we talked about the number $e$, we proved that the sequence ${x}_{{n}}={{\left({1}+\frac{{1}}{{n}}\right)}}^{{n}}$ has limit $e$.

Example 11. Prove that $\lim_{{{h}\to{0}}}\frac{{{{a}}^{{h}}-{1}}}{{h}}={\ln{{\left({a}\right)}}}$.

Let ${{a}}^{{h}}-{1}={x}$ then ${x}\to{0}$ as ${h}\to{0}$. And ${h}={\log}_{{a}}{\left({x}+{1}\right)}=\frac{{\ln{{\left({x}+{1}\right)}}}}{{\ln{{\left({a}\right)}}}}$

So, $\lim_{{{h}\to{0}}}\frac{{{{a}}^{{h}}-{1}}}{{h}}=\lim_{{{x}\to{0}}}\frac{{x}}{{\frac{{{\ln{{\left({x}+{1}\right)}}}}}{{{\ln{{\left({a}\right)}}}}}}}={\ln{{\left({a}\right)}}}\lim_{{{x}\to{0}}}\frac{{x}}{{{\ln{{\left({x}+{1}\right)}}}}}={\ln{{\left({a}\right)}}}\lim_{{{x}\to{0}}}\frac{{1}}{{\frac{{1}}{{x}}{\ln{{\left({x}+{1}\right)}}}}}=$

$={\ln{{\left({a}\right)}}}\lim_{{{x}\to{0}}}\frac{{1}}{{{{\ln{{\left({1}+{x}\right)}}}}^{{\frac{{1}}{{x}}}}}}={\ln{{\left({a}\right)}}}\frac{{1}}{{{\ln{{\left(\lim_{{{x}\to{0}}}{{\left({1}+{x}\right)}}^{{\frac{{1}}{{x}}}}\right)}}}}}={\ln{{\left({a}\right)}}}\frac{{1}}{{{\ln{{\left({e}\right)}}}}}={\ln{{\left({a}\right)}}}\frac{{1}}{{1}}={\ln{{\left({a}\right)}}}$

Example 12. Find $\lim_{{{x}\to{0}}}{{x}}^{{x}}$.

Since $\lim_{{{x}\to{0}}}{x}={0}$ then we have case b) here.

Let ${B}={{x}}^{{x}}$ then ${\ln{{\left({B}\right)}}}={\ln{{\left({{x}}^{{x}}\right)}}}={x}{\ln{{\left({x}\right)}}}$.

It is known that $\lim_{{{x}\to{0}}}{x}{\ln{{\left({x}\right)}}}={0}$ that's why ${\ln{{\left({B}\right)}}}={0}$ or ${B}={1}$.

Thus, $\lim_{{{x}\to{0}}}{{x}}^{{x}}={1}$.