Limit Laws and Algebraic Techniques

Both given limits exist and are finite, so the Constant Multiple, Difference and Sum Laws apply to the top and bottom separately.

Numerator.

$\lim_{x\to 2}\left(2f(x)-g(x)\right)=2\lim_{x\to 2}f(x)-\lim_{x\to 2}g(x)=2(5)-(-2)=10+2=12$

Denominator.

$\lim_{x\to 2}\left(f(x)+g(x)\right)=\lim_{x\to 2}f(x)+\lim_{x\to 2}g(x)=5+(-2)=3$

Check the hypothesis. The Quotient Law requires the limit of the denominator to be nonzero. Here it is $3\ne 0$, so the law applies:

$\lim_{x\to 2}\frac{2f(x)-g(x)}{f(x)+g(x)}=\frac{12}{3}=4$

Answer: $4$. The hypothesis that must be checked is $\lim_{x\to 2}\left(f(x)+g(x)\right)\ne 0$. Had it been $0$, the Quotient Law would say nothing and the limit could be finite, infinite, or nonexistent.

Step 1 — try substitution. At $x=-1$ the numerator is $1-3+2=0$ and the denominator is $1-1=0$. The form is $\frac00$, so the Quotient Law does not apply and more work is needed.

Step 2 — factor. Both quadratics vanish at $x=-1$, so $(x+1)$ is a factor of each:

$x^{2}+3x+2=(x+1)(x+2),\qquad x^{2}-1=(x+1)(x-1)$

Step 3 — cancel. For $x\ne -1$ the factor $x+1$ is a nonzero number, so it may be cancelled:

$\frac{x^{2}+3x+2}{x^{2}-1}=\frac{(x+1)(x+2)}{(x+1)(x-1)}=\frac{x+2}{x-1}\qquad (x\ne -1)$

Step 4 — substitute into the simplified form. $\frac{x+2}{x-1}$ is a rational function whose denominator at $x=-1$ is $-2\ne 0$, so it is continuous there and direct substitution is valid:

$\lim_{x\to -1}\frac{x+2}{x-1}=\frac{-1+2}{-1-1}=\frac{1}{-2}=-\frac{1}{2}$

Answer: $-\dfrac{1}{2}$.

Step 1 — try substitution. Numerator $\to\sqrt{9}-3=0$, denominator $\to 0$. Indeterminate $\frac00$.

Step 2 — rationalise the numerator. Multiply by the conjugate $\sqrt{9+x}+3$ over itself (this is legal since $\sqrt{9+x}+3\ge 3>0$ for $x$ near $0$):

Step 3 — cancel. For $x\ne 0$:

$\frac{x}{x\left(\sqrt{9+x}+3\right)}=\frac{1}{\sqrt{9+x}+3}$

Step 4 — substitute. The function $\frac{1}{\sqrt{9+x}+3}$ is continuous at $x=0$ (the root is defined and the denominator is $6\ne 0$):

$\lim_{x\to 0}\frac{1}{\sqrt{9+x}+3}=\frac{1}{\sqrt{9}+3}=\frac{1}{6}$

Answer: $\dfrac{1}{6}$. (Sanity check: at $x=0.001$ the original quotient is $0.166662$..., close to $\frac16\approx 0.1667$.)

Step 1 — check the form. As $x\to 0$ both $\sin 7x\to 0$ and $\sin 3x\to 0$, so this is $\frac00$; the Quotient Law does not apply directly.

Step 2 — insert the matching arguments. For $x\ne 0$ small enough that $\sin 3x\ne 0$, multiply and divide to build the standard ratios:

Step 3 — apply the standard limit via substitution. Put $u=7x$: as $x\to 0$ with $x\ne 0$ we get $u\to 0$ with $u\ne 0$, so $\lim_{x\to 0}\frac{\sin 7x}{7x}=\lim_{u\to 0}\frac{\sin u}{u}=1$. Likewise with $v=3x$, $\lim_{x\to 0}\frac{\sin 3x}{3x}=1$, and since that limit is nonzero the Quotient Law gives $\lim_{x\to 0}\frac{3x}{\sin 3x}=\frac{1}{1}=1$.

Step 4 — combine. All three factors have limits, so the Product Law applies:

$\lim_{x\to 0}\frac{\sin 7x}{\sin 3x}=\frac{7}{3}\cdot 1\cdot 1=\frac{7}{3}$

Answer: $\dfrac{7}{3}$. (Numerical check: at $x=0.001$ the quotient is $2.33332$, heading to $\frac{7}{3}=2.3333$...)

Step 1 — recognise the form. Each term blows up as $t\to 0$, so the Difference Law does not apply (neither piece has a limit). Combine first.

Step 2 — common denominator. For $t\ne 0$ and $t>-1$:

$\frac{1}{t\sqrt{1+t}}-\frac{1}{t}=\frac{1-\sqrt{1+t}}{t\sqrt{1+t}}$

This is now $\frac00$.

Step 3 — rationalise the numerator with the conjugate $1+\sqrt{1+t}$:

Step 4 — cancel the factor $t\ne 0$:

$\frac{-t}{t\sqrt{1+t}\left(1+\sqrt{1+t}\right)}=\frac{-1}{\sqrt{1+t}\left(1+\sqrt{1+t}\right)}$

Step 5 — substitute. This is continuous at $t=0$ and its denominator there is $1\cdot(1+1)=2\ne 0$:

$\lim_{t\to 0}\frac{-1}{\sqrt{1+t}\left(1+\sqrt{1+t}\right)}=\frac{-1}{1\cdot 2}=-\frac{1}{2}$

Answer: $-\dfrac{1}{2}$. (Check at $t=0.001$: $999.5004-1000=-0.4996$, consistent.)

Step 1 — name the bounds. Let $g(x)=4x-9$ and $h(x)=x^{2}-4x+7$, so $g(x)\le f(x)\le h(x)$.

Step 2 — check the interval hypothesis. The Squeeze Theorem needs the inequality to hold on an open interval containing $a=4$, except possibly at $a$. The inequality is given for all $x\ge 0$, and the open interval $(3,5)$ is contained in $[0,\infty)$ and contains $4$. Hypothesis satisfied.

Step 3 — compute the outer limits. Both $g$ and $h$ are polynomials, hence continuous everywhere, so direct substitution applies:

$\lim_{x\to 4}g(x)=4(4)-9=16-9=7$

$\lim_{x\to 4}h(x)=4^{2}-4(4)+7=16-16+7=7$

Step 4 — the outer limits agree. Both equal $L=7$, which is the remaining hypothesis.

Step 5 — conclude. By the Squeeze Theorem, $\lim_{x\to 4}f(x)$ exists and

$\lim_{x\to 4}f(x)=7$

Answer: $7$. Note that we never needed a formula for $f$, and $f$ need not be continuous at $4$ — the squeeze forces the limit regardless. Note also that the argument works only at $x=4$: at $x=3$, for instance, $g(3)=3$ and $h(3)=4$ differ, so the theorem would give no conclusion there.

Step 1 — check the form. $1-\cos x\to 0$ and $x\sin x\to 0$, so this is $\frac00$.

Step 2 — multiply by the conjugate $1+\cos x$. For $x$ near $0$ with $x\ne 0$ we have $\sin x\ne 0$ and $1+\cos x\ne 0$, so:

using the Pythagorean identity $1-\cos^{2}x=\sin^{2}x$ and cancelling one factor of $\sin x\ne 0$.

Step 3 — take limits factor by factor. $\lim_{x\to 0}\frac{\sin x}{x}=1$ (standard limit), and $\frac{1}{1+\cos x}$ is continuous at $0$ with value $\frac{1}{1+1}=\frac12$. Both limits exist, so the Product Law applies:

$\lim_{x\to 0}\frac{1-\cos x}{x\sin x}=1\cdot\frac{1}{2}=\frac{1}{2}$

Answer: $\dfrac{1}{2}$.

Alternative route: write $\frac{1-\cos x}{x\sin x}=\frac{1-\cos x}{x^{2}}\cdot\frac{x}{\sin x}\to\frac12\cdot 1=\frac12$, using the standard result $\lim_{x\to 0}\frac{1-\cos x}{x^{2}}=\frac12$. Same answer, which is a good independent check.

Step 1 — check the form. At $x=1$ both $x^{1/3}-1$ and $x^{1/2}-1$ equal $0$: indeterminate $\frac00$.

Step 2 — change variable. Let $u=x^{1/6}$, so $x=u^{6}$ with $u>0$, and then $x^{1/3}=u^{2}$, $x^{1/2}=u^{3}$.

Hypotheses for the substitution: $u=x^{1/6}$ is continuous on $x>0$ with $u\to 1$ as $x\to 1$, and $u\ne 1$ whenever $x\ne 1$ (the sixth-root function is strictly increasing on $x>0$, hence one-to-one). So the change of variable is valid and

$\lim_{x\to 1}\frac{x^{1/3}-1}{x^{1/2}-1}=\lim_{u\to 1}\frac{u^{2}-1}{u^{3}-1}$

Step 3 — factor. Using $u^{2}-1=(u-1)(u+1)$ and $u^{3}-1=(u-1)\left(u^{2}+u+1\right)$:

$\frac{u^{2}-1}{u^{3}-1}=\frac{(u-1)(u+1)}{(u-1)\left(u^{2}+u+1\right)}=\frac{u+1}{u^{2}+u+1}\qquad (u\ne 1)$

Step 4 — substitute. The simplified expression is a rational function whose denominator at $u=1$ is $1+1+1=3\ne 0$, so it is continuous there:

$\lim_{u\to 1}\frac{u+1}{u^{2}+u+1}=\frac{2}{3}$

Answer: $\dfrac{2}{3}$. (Check at $x=1.0001$: the original quotient is $0.66666$...)

Step 1 — check the form. As $x\to 0$, $(1+x)^{1/3}\to 1$, so the numerator $\to 0$ and the denominator $\to 0$: indeterminate $\frac00$.

Step 2 — set up the cube-root conjugate. Write $a=(1+x)^{1/3}$, so $a^{3}=1+x$, and $b=1$. The identity is

$(a-b)\left(a^{2}+ab+b^{2}\right)=a^{3}-b^{3}$

so multiply top and bottom by $a^{2}+a+1$:

This multiplication is legitimate because $a^{2}+a+1\ne 0$ for real $a$ (its discriminant is $1-4=-3<0$), and in fact $a^{2}+a+1\to 3$ near $x=0$.

Step 3 — cancel the factor $x\ne 0$:

$\frac{x}{x\left(a^{2}+a+1\right)}=\frac{1}{a^{2}+a+1}=\frac{1}{(1+x)^{2/3}+(1+x)^{1/3}+1}$

Step 4 — substitute. The cube root is continuous everywhere, so the denominator is continuous at $x=0$ with value $1+1+1=3\ne 0$:

$\lim_{x\to 0}\frac{(1+x)^{1/3}-1}{x}=\frac{1}{3}$

Answer: $\dfrac{1}{3}$. (Check at $x=0.001$: $\frac{1.000333222-1}{0.001}=0.33322$..., heading to $\frac13$. This limit is also the derivative of $(1+x)^{1/3}$ at $x=0$, and $\frac{d}{dx}(1+x)^{1/3}=\frac13(1+x)^{-2/3}$ equals $\frac13$ there — an independent confirmation.)

Step 1 — check the form. $1-\cos 4x\to 0$ and $x\tan 2x\to 0$: indeterminate $\frac00$.

Step 2 — remove the tangent and the cosine. For $x\ne 0$ small (so $\sin 2x\ne 0$, $\cos 2x\ne 0$, $1+\cos 4x\ne 0$):