Math Analysis 1

Solved limit exercises, from basic to exam level

Twenty-two exercises with solutions: standard limits, indeterminate forms, 1 to the infinity, squeezing, Taylor expansions, parameters, sequences and asymptotes, with graphs.

ielettronica 25 October 2026 12 min read 0 comments

Exercises on limits of functions and sequences of the kind set in Calculus 1 (Mathematical Analysis 1) courses, in three levels: basic, for the direct use of standard limits and elementary techniques; intermediate, with $1^\infty$ and $0\cdot\infty$ forms, changes of variable and squeezing; exam, with Taylor expansions, parameters and asymptotes, as in written exams and mid-term tests. The result is given straight away; the full solution opens with a click. The tables of standard limits and expansions are in the in-depth study linked at the top.

Basic level

Exercise 1. A standard limit with sine

Compute $\displaystyle\lim_{x\to 0}\frac{\sin 5x}{3x}$ .

Result: $\frac{5}{3}$ .

Solution

As $x \to 0$ also $5x \to 0$ , so $\sin 5x \sim 5x$ . Substituting in the quotient: $\dfrac{5x}{3x} = \dfrac{5}{3}$ .

Alternatively, without equivalences: $\dfrac{\sin 5x}{3x} = \dfrac{5}{3}\cdot\dfrac{\sin 5x}{5x}$ , and the second factor tends to 1.

Graph for exercise 1 — The blue curve is sine of 5 x over 3 x; as x tends to 0 it approaches the dashed red line, the value of the limit: 5/3. The empty circle shows that the function is not defined at that point.

Exercise 2. Cosine and sine together

Compute $\displaystyle\lim_{x\to 0}\frac{1 - \cos x}{x\sin x}$ .

Result: $\frac{1}{2}$ .

Solution

Numerator and denominator are products and quotients of infinitesimals: they can be replaced by equivalents. $1 - \cos x \sim \frac{x^2}{2}$ and $x\sin x \sim x\cdot x = x^2$ . The limit is $\dfrac{x^2/2}{x^2} = \dfrac{1}{2}$ .

Graph for exercise 2 — The blue curve is 1 minus cosine of x over x times sine of x; as x tends to 0 it approaches the dashed red line, the value of the limit: 1/2. The empty circle shows that the function is not defined at that point.

Exercise 3. Ratio of polynomials at infinity

Compute $\displaystyle\lim_{x\to+\infty}\frac{3x^2 - x + 1}{2x^2 + 5}$ .

Result: $\frac{3}{2}$ .

Solution

Form $\infty/\infty$ . Factor out the highest power above and below:

$\frac{x^2\left(3 - \frac{1}{x} + \frac{1}{x^2}\right)}{x^2\left(2 + \frac{5}{x^2}\right)} \to \frac{3}{2}$

General rule for ratios of polynomials at infinity: only the ratio of the highest-degree terms matters. With equal degrees you get the ratio of the coefficients; with a higher-degree numerator $\pm\infty$ ; with a higher-degree denominator 0.

Graph for exercise 3 — The blue curve is the ratio of the two polynomials; as x tends to plus infinity it approaches the dashed red line, the value of the limit: 3/2.

Exercise 4. A difference of infinities with a square root

Compute $\displaystyle\lim_{x\to+\infty}\left(\sqrt{x^2 + x} - x\right)$ .

Result: $\frac{1}{2}$ .

Solution

Form $\infty - \infty$ . Rationalise by multiplying and dividing by $\sqrt{x^2 + x} + x$ :

$\sqrt{x^2 + x} - x = \frac{(x^2 + x) - x^2}{\sqrt{x^2 + x} + x} = \frac{x}{x\left(\sqrt{1 + \frac{1}{x}} + 1\right)} = \frac{1}{\sqrt{1 + \frac{1}{x}} + 1} \to \frac{1}{2}$

(For $x \gt 0$ , $\sqrt{x^2} = x$ ; as $x \to -\infty$ it would be $\sqrt{x^2} = -x$ and the limit would change: it becomes $+\infty$ .)

Graph for exercise 4 — The blue curve is square root of x squared plus x, minus x; as x tends to plus infinity it approaches the dashed red line, the value of the limit: 1/2.

Exercise 5. Exponential and logarithm

Compute $\displaystyle\lim_{x\to 0}\frac{e^{2x} - 1}{\ln(1 + 3x)}$ .

Result: $\frac{2}{3}$ .

Solution

$e^{2x} - 1 \sim 2x$ and $\ln(1 + 3x) \sim 3x$ . The limit is $\dfrac{2x}{3x} = \dfrac{2}{3}$ .

Graph for exercise 5 — The blue curve is e to the 2 x minus 1, over log of 1 plus 3 x; as x tends to 0 it approaches the dashed red line, the value of the limit: 2/3. The empty circle shows that the function is not defined at that point.

Exercise 6. Factor and simplify

Compute $\displaystyle\lim_{x\to 2}\frac{x^2 - 4}{x^2 - 3x + 2}$ .

Result: 4.

Solution

Substituting $x = 2$ gives $0/0$ : both polynomials vanish at 2, so both are divisible by $x - 2$ (factor theorem).

$\frac{x^2 - 4}{x^2 - 3x + 2} = \frac{(x - 2)(x + 2)}{(x - 2)(x - 1)} = \frac{x + 2}{x - 1} \to \frac{4}{1} = 4$

Graph for exercise 6 — The blue curve is the ratio of the two polynomials, which has a vertical asymptote at x equal to 1; as x tends to 2 it approaches the dashed red line, the value of the limit: 4. The empty circle shows that the function is not defined at that point.

Intermediate level

Exercise 7. A 1 to the infinity form

Compute $\displaystyle\lim_{x\to+\infty}\left(\frac{x - 1}{x + 2}\right)^x$ .

Result: $e^{-3}$ .

Solution

The base tends to 1 and the exponent to $+\infty$ : form $1^\infty$ . Write the base as 1 plus an infinitesimal: $\dfrac{x - 1}{x + 2} = 1 - \dfrac{3}{x + 2}$ .

Switch to the exponential: $\left(\dfrac{x - 1}{x + 2}\right)^x = e^{x\ln\left(1 - \frac{3}{x + 2}\right)}$ . Since $\ln(1 + u) \sim u$ as $u \to 0$ :

$x\ln\left(1 - \frac{3}{x + 2}\right) \sim x\cdot\left(-\frac{3}{x + 2}\right) \to -3$

and the limit is $e^{-3}$ . The same result follows from the corollary $\left(1 + \frac{b}{x}\right)^x \to e^b$ , after noting that $x + 2 \sim x$ .

Graph for exercise 7 — The blue curve is x minus 1 over x plus 2, to the power x; as x tends to plus infinity it approaches the dashed red line, the value of the limit: e to the minus 3, about 0.05.

Exercise 8. When equivalences are not enough

Compute $\displaystyle\lim_{x\to 0}\frac{x - \sin x}{x^3}$ .

Result: $\frac{1}{6}$ .

Solution

Substituting $\sin x \sim x$ in the numerator would give $x - x = 0$ : wrong, because in a difference the leading terms cancel. The next term of the expansion is needed: $\sin x = x - \frac{x^3}{6} + o(x^4)$ .

$\frac{x - \sin x}{x^3} = \frac{\frac{x^3}{6} + o(x^4)}{x^3} = \frac{1}{6} + o(x) \to \frac{1}{6}$

With l’Hôpital’s rule (allowed: form $0/0$ ): $\dfrac{1 - \cos x}{3x^2} \to \dfrac{1/2}{3} = \dfrac{1}{6}$ .

Graph for exercise 8 — The blue curve is x minus sine of x, over x cubed; as x tends to 0 it approaches the dashed red line, the value of the limit: 1/6. The empty circle shows that the function is not defined at that point.

Exercise 9. The logarithm of cosine

Compute $\displaystyle\lim_{x\to 0}\frac{\ln(\cos x)}{x^2}$ .

Result: $-\frac{1}{2}$ .

Solution

Write $\cos x = 1 + (\cos x - 1)$ , with $u = \cos x - 1 \to 0$ . Then $\ln(\cos x) = \ln(1 + u) \sim u = \cos x - 1 \sim -\dfrac{x^2}{2}$ .

The limit is $\dfrac{-x^2/2}{x^2} = -\dfrac{1}{2}$ . This step comes up often: $\ln f \sim f - 1$ when $f \to 1$ .

Graph for exercise 9 — The blue curve is log of cosine of x over x squared; as x tends to 0 it approaches the dashed red line, the value of the limit: minus 1/2. The empty circle shows that the function is not defined at that point.

Exercise 10. Zero times infinity

Compute $\displaystyle\lim_{x\to 0^+}x\ln x$ .

Result: 0.

Solution

Form $0\cdot(-\infty)$ . Set $x = 1/t$ , with $t \to +\infty$ : $x\ln x = \dfrac{1}{t}\ln\dfrac{1}{t} = -\dfrac{\ln t}{t} \to 0$ , by the hierarchy of infinities (the logarithm is negligible compared with any power).

More generally $x^\alpha\ln x \to 0$ as $x \to 0^+$ for every $\alpha \gt 0$ : the power “wins” near zero too.

Graphs of x times log x and of x to the x — The orange curve, x log x, tends to zero as x tends to zero from the right and has a minimum of minus 1 over e at x equal to 1 over e. The blue curve, x to the x, which is e raised to the orange curve, tends to 1 and has its minimum at x equal to 1 over e.

Exercise 11. Zero to the zero

Compute $\displaystyle\lim_{x\to 0^+}x^x$ .

Result: 1.

Solution

Form $0^0$ . Switch to the exponential: $x^x = e^{x\ln x}$ . The exponent tends to 0 (previous exercise) and the exponential is continuous, so $x^x \to e^0 = 1$ .

Exercise 12. A change of variable at infinity

Compute $\displaystyle\lim_{x\to+\infty}x\left(e^{1/x} - 1\right)$ .

Result: 1.

Solution

Form $\infty\cdot 0$ . With $t = 1/x \to 0^+$ : $x\left(e^{1/x} - 1\right) = \dfrac{e^t - 1}{t} \to 1$ .

This is an example of the rule: when $x \to \infty$ , the change $t = 1/x$ brings the standard limits back “to 0”.

Graph for exercise 12 — The blue curve is x times e to the 1 over x minus 1; as x tends to plus infinity it approaches the dashed red line, the value of the limit: 1.

Exercise 13. Moving to the right point

Compute $\displaystyle\lim_{x\to\pi/2}\left(\frac{\pi}{2} - x\right)\tan x$ .

Result: 1.

Solution

The standard limits are “at 0”: set $t = \frac{\pi}{2} - x \to 0$ . Then $\tan x = \tan\left(\frac{\pi}{2} - t\right) = \dfrac{\cos t}{\sin t}$ (complementary-angle formulas).

$\left(\frac{\pi}{2} - x\right)\tan x = t\cdot\frac{\cos t}{\sin t} = \frac{t}{\sin t}\cdot\cos t \to 1\cdot 1 = 1$

The limit is the same from the right and from the left: for $x \gt \pi/2$ both $t$ and $\tan x$ are negative.

Graph for exercise 13 — The blue curve is pi over 2 minus x, times tangent of x; as x tends to pi over 2 it approaches the dashed red line, the value of the limit: 1. The empty circle shows that the function is not defined at that point.

Exercise 14. A ratio of logarithms

Compute $\displaystyle\lim_{x\to+\infty}\frac{\ln(x^2 + 1)}{\ln x}$ .

Result: 2.

Solution

Factor $x^2$ out inside the logarithm: $\ln(x^2 + 1) = \ln\left[x^2\left(1 + \frac{1}{x^2}\right)\right] = 2\ln x + \ln\left(1 + \frac{1}{x^2}\right)$ .

The second term tends to 0, so $\dfrac{\ln(x^2 + 1)}{\ln x} = 2 + \dfrac{\ln(1 + 1/x^2)}{\ln x} \to 2$ .

Graph for exercise 14 — The blue curve is log of x squared plus 1 over log of x; as x tends to plus infinity it approaches the dashed red line, the value of the limit: 2.

Exercise 15. The squeeze theorem

Compute $\displaystyle\lim_{x\to 0}x^2\sin\frac{1}{x}$ .

Result: 0.

Solution

$\sin(1/x)$ has no limit as $x \to 0$ : it oscillates between $-1$ and 1 faster and faster. But it is bounded: $\left|\sin\frac{1}{x}\right| \le 1$ , so

$-x^2 \le x^2\sin\frac{1}{x} \le x^2$

Both bounds tend to 0, and by the squeeze theorem so does the middle function. Useful rule: an infinitesimal times a bounded function is an infinitesimal.

The function x squared times sine of 1 over x squeezed between two parabolas — The blue curve oscillates faster and faster near the origin, but stays between the parabolas minus x squared and x squared, in orange, which both tend to zero and force it to do the same.

Exam level

Exercise 16. Taylor with a square root

Compute $\displaystyle\lim_{x\to 0}\frac{\sqrt{1 + 2x} - 1 - x}{x^2}$ .

Result: $-\frac{1}{2}$ .

Solution

The equivalence $\sqrt{1 + 2x} - 1 \sim x$ would cancel the $-x$ and give 0: a mistake. Expand to second order $(1 + u)^{1/2} = 1 + \frac{u}{2} - \frac{u^2}{8} + o(u^2)$ with $u = 2x$ :

$\sqrt{1 + 2x} = 1 + x - \frac{4x^2}{8} + o(x^2) = 1 + x - \frac{x^2}{2} + o(x^2)$

Numerator: $-\dfrac{x^2}{2} + o(x^2)$ . The limit is $-\dfrac{1}{2}$ .

Graph for exercise 16 — The blue curve is square root of 1 plus 2 x, minus 1 minus x, over x squared; as x tends to 0 it approaches the dashed red line, the value of the limit: minus 1/2. The empty circle shows that the function is not defined at that point.

Exercise 17. Sine minus x times cosine

Compute $\displaystyle\lim_{x\to 0}\frac{\sin x - x\cos x}{x^3}$ .

Result: $\frac{1}{3}$ .

Solution

The denominator is of order 3, so expand the numerator up to order 3:

$\sin x = x - \frac{x^3}{6} + o(x^3)$ ; $x\cos x = x\left(1 - \frac{x^2}{2} + o(x^2)\right) = x - \frac{x^3}{2} + o(x^3)$ .

Difference: $-\frac{x^3}{6} + \frac{x^3}{2} + o(x^3) = \frac{x^3}{3} + o(x^3)$ . The limit is $\dfrac{1}{3}$ .

Practical rule: expand each term up to the order of the denominator (or up to the first order that does not cancel).

Graph for exercise 17 — The blue curve is sine of x minus x cosine of x, over x cubed; as x tends to 0 it approaches the dashed red line, the value of the limit: 1/3. The empty circle shows that the function is not defined at that point.

Exercise 18. Difference of two infinities with Taylor

Compute $\displaystyle\lim_{x\to 0}\left(\frac{1}{\sin^2 x} - \frac{1}{x^2}\right)$ .

Result: $\frac{1}{3}$ .

Solution

Form $\infty - \infty$ . Common denominator:

$\frac{1}{\sin^2 x} - \frac{1}{x^2} = \frac{x^2 - \sin^2 x}{x^2\sin^2 x}$

The denominator is $\sim x^2\cdot x^2 = x^4$ . In the numerator: $\sin^2 x = \left(x - \frac{x^3}{6} + o(x^3)\right)^2 = x^2 - \frac{x^4}{3} + o(x^4)$ (the cross term is $2\cdot x\cdot\left(-\frac{x^3}{6}\right) = -\frac{x^4}{3}$ ).

Numerator: $\frac{x^4}{3} + o(x^4)$ . The limit is $\dfrac{x^4/3}{x^4} = \dfrac{1}{3}$ .

Graph of 1 over sine squared of x minus 1 over x squared — The two separate functions blow up at zero, but their difference stays bounded and tends to one third, shown by the dashed line.

Exercise 19. A 1 to the infinity form with cosine

Compute $\displaystyle\lim_{x\to 0}(\cos x)^{1/x^2}$ .

Result: $e^{-1/2} = \dfrac{1}{\sqrt e} \approx 0.607$ .

Solution

$(\cos x)^{1/x^2} = e^{\ln(\cos x)/x^2}$ . The exponent tends to $-\frac{1}{2}$ (exercise 9), so the limit is $e^{-1/2}$ .

Graph for exercise 19 — The blue curve is cosine of x to the power 1 over x squared; as x tends to 0 it approaches the dashed red line, the value of the limit: 1 over root e, about 0.607. The empty circle shows that the function is not defined at that point.

Exercise 20. Finding the parameters

Find the real numbers $a$ and $b$ for which the limit $\displaystyle\lim_{x\to 0}\frac{\ln(1 + x) - ax - bx^2}{x^3}$ is finite, and compute it.

Result: $a = 1$ , $b = -\frac{1}{2}$ ; the limit is $\frac{1}{3}$ .

Solution

Expansion: $\ln(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} + o(x^3)$ . The numerator is

$(1 - a)x + \left(-\frac{1}{2} - b\right)x^2 + \frac{x^3}{3} + o(x^3)$

Divided by $x^3$ : $\dfrac{1 - a}{x^2} + \dfrac{-\frac{1}{2} - b}{x} + \dfrac{1}{3} + o(1)$ .

1. If $a \ne 1$ , the first term dominates and the limit is infinite.

2. If $a = 1$ but $b \ne -\frac{1}{2}$ , the second term dominates: it tends to $+\infty$ on one side and to $-\infty$ on the other, so the limit does not exist.

3. With $a = 1$ and $b = -\frac{1}{2}$ what remains is $\frac{1}{3} + o(1)$ : the limit is $\frac{1}{3}$ .

Meaning: $x - \frac{x^2}{2}$ is the second-order Taylor polynomial of the logarithm, the only one that approximates $\ln(1 + x)$ with an error of order $x^3$ .

Graph for exercise 20 — The blue curve is the ratio with a equal to 1 and b equal to minus 1/2; as x tends to 0 it approaches the dashed red line, the value of the limit: 1/3. The empty circle shows that the function is not defined at that point.

Exercise 21. The limit of a sequence

Compute $\displaystyle\lim_{n\to\infty}n\left(\sqrt[n]{2} - 1\right)$ .

Result: $\ln 2 \approx 0.693$ .

Solution

Form $\infty\cdot 0$ . Write $\sqrt[n]{2} = 2^{1/n}$ and set $t = 1/n \to 0^+$ :

$n\left(2^{1/n} - 1\right) = \frac{2^t - 1}{t} \to \ln 2$

Passing from the sequence to the function is allowed by the sequential characterisation of limits (lesson on sequences): if the limit of the function exists, the sequence has the same limit.

Graph for exercise 21 — The blue curve is the terms of the sequence n times the n-th root of 2 minus 1; as n grows it approaches the dashed red line, the value of the limit: log 2, about 0.693.

Exercise 22. Asymptotes of a rational function

Find the asymptotes of $f(x) = \dfrac{x^2 + 1}{x - 1}$ and determine on which side of the graph the oblique asymptote lies.

Result: vertical asymptote $x = 1$ ; oblique asymptote $y = x + 1$ , with the graph above the asymptote for $x \gt 1$ and below it for $x \lt 1$ .

Solution

Vertical. The domain is $x \ne 1$ . As $x \to 1$ the numerator tends to 2 and the denominator to 0: $f \to +\infty$ as $x \to 1^+$ and $f \to -\infty$ as $x \to 1^-$ . So $x = 1$ is a vertical asymptote.

Horizontal. As $x \to \pm\infty$ , $f(x) \sim x \to \pm\infty$ : there is none.

Oblique. $m = \lim\dfrac{f(x)}{x} = \lim\dfrac{x^2 + 1}{x^2 - x} = 1$ ; $q = \lim\left(f(x) - x\right) = \lim\dfrac{x^2 + 1 - x^2 + x}{x - 1} = \lim\dfrac{x + 1}{x - 1} = 1$ . Asymptote $y = x + 1$ , both at $+\infty$ and at $-\infty$ .

Position. By polynomial division: $x^2 + 1 = (x - 1)(x + 1) + 2$ , so $f(x) = x + 1 + \dfrac{2}{x - 1}$ . The difference $f(x) - (x + 1) = \dfrac{2}{x - 1}$ is positive for $x \gt 1$ (graph above the asymptote) and negative for $x \lt 1$ (below).

For rational functions whose numerator degree is the denominator degree plus one, division gives the oblique asymptote and the position straight away, without computing limits.

Graph of x squared plus 1 over x minus 1 with its asymptotes — The vertical asymptote x equal to 1 and the oblique asymptote y equal to x plus 1 are dashed. The right branch lies above the oblique asymptote and has a minimum, the left one lies below and has a maximum.

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Basic level

Exercise 1. A standard limit with sine

Exercise 2. Cosine and sine together

Exercise 3. Ratio of polynomials at infinity

Exercise 4. A difference of infinities with a square root

Exercise 5. Exponential and logarithm

Exercise 6. Factor and simplify

Intermediate level

Exercise 7. A 1 to the infinity form

Exercise 8. When equivalences are not enough

Exercise 9. The logarithm of cosine

Exercise 10. Zero times infinity

Exercise 11. Zero to the zero

Exercise 12. A change of variable at infinity

Exercise 13. Moving to the right point

Exercise 14. A ratio of logarithms

Exercise 15. The squeeze theorem

Exam level

Exercise 16. Taylor with a square root

Exercise 17. Sine minus x times cosine

Exercise 18. Difference of two infinities with Taylor

Exercise 19. A 1 to the infinity form with cosine

Exercise 20. Finding the parameters

Exercise 21. The limit of a sequence

Exercise 22. Asymptotes of a rational function

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