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3"},"content":{"rendered":""},"excerpt":{"rendered":""},"featured_media":0,"featured_media_url":"$undefined","acf":{"duree_estimee":10,"niveau_difficulte":"Intermediate","categorie":"Limits & Continuity","nombre_questions":10,"score_minimum":70,"ordre_questions":"Fixe","questions":[{"id":"cmpz9r9xc07sajgllhehnq934","texte_question":"

$$\\lim_{x\\to3}\\frac{x^3-27}{x^2-9}=?$$

","type_question":"QCM","explication":"

Substitute $$x=3$$ to check the form:

$$\\frac{3^3-27}{3^2-9}=\\frac{27-27}{9-9}=\\frac{0}{0}$$

This is an indeterminate form, so factor numerator and denominator.

Factor the numerator as a difference of cubes:

$$x^3-27=(x-3)(x^2+3x+9)$$

Factor the denominator as a difference of squares:

$$x^2-9=(x-3)(x+3)$$

Rewrite the expression:

$$\\frac{(x-3)(x^2+3x+9)}{(x-3)(x+3)}$$

Cancel the common factor $$x-3$$:

$$\\frac{x^2+3x+9}{x+3}$$

Now substitute $$x=3$$:

$$\\frac{3^2+3(3)+9}{3+3}=\\frac{9+9+9}{6}=\\frac{27}{6}$$

Simplify:

$$\\frac{27}{6}=\\frac{9}{2}$$

Final answer: $$\\frac{9}{2}$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$6$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$\\frac{7}{2}$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$3$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$\\frac{9}{2}$$","correcte":true,"explication":"","imageUrl":"$undefined"}]},{"id":"cmpz9r9xg07sgjgll4lfeby39","texte_question":"

$$\\lim_{x\\to\\infty}\\frac{7-x^2}{5x^2+2x-4}=?$$

","type_question":"QCM","explication":"

The highest power in the numerator is $$x^2$$ with leading term $$-x^2$$.

The highest power in the denominator is $$x^2$$ with leading term $$5x^2$$.

As $$x\\to\\infty$$, the limit equals the ratio of the leading coefficients:

$$\\lim_{x\\to\\infty}\\frac{7-x^2}{5x^2+2x-4}=\\frac{-1}{5}$$

Final answer: $$-\\frac{1}{5}$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$-\\frac{1}{5}$$","correcte":true,"explication":"","imageUrl":"$undefined"},{"texte":"$$$-\\frac{2}{5}$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$\\frac{1}{5}$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$-\\frac{1}{3}$$","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmpz9r9xj07smjgllpnd2sqdf","texte_question":"

$$\\lim_{x\\to-\\infty}\\frac{4x^3-32}{15x^2+6x+1}=?$$

","type_question":"QCM","explication":"

The highest power in the numerator is $$x^3$$.

The highest power in the denominator is $$x^2$$.

Since the numerator has higher degree than the denominator, the expression grows without bound in magnitude.

Divide numerator and denominator by $$x^2$$:

$$\\frac{4x^3-32}{15x^2+6x+1}=\\frac{\\frac{4x^3}{x^2}-\\frac{32}{x^2}}{\\frac{15x^2}{x^2}+\\frac{6x}{x^2}+\\frac{1}{x^2}}$$

Simplify:

$$=\\frac{4x-\\frac{32}{x^2}}{15+\\frac{6}{x}+\\frac{1}{x^2}}$$

As $$x\\to-\\infty$$:

$$-\\frac{32}{x^2}\\to0$$

$$\\frac{6}{x}\\to0$$

$$\\frac{1}{x^2}\\to0$$

So, the expression behaves like:

$$\\frac{4x}{15}$$

Since $$x\\to-\\infty$$:

$$\\frac{4x}{15}\\to-\\infty$$

Final answer: The limit diverges to $$-\\infty$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$-5$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"The limit diverges to $$-\\infty$$","correcte":true,"explication":"","imageUrl":"$undefined"},{"texte":"$$$\\infty$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$0$$","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmpz9r9xl07ssjgllgj9fqiok","texte_question":"

$$\\lim_{x\\to2}\\frac{5x^2+40}{x^3-8}=?$$

","type_question":"QCM","explication":"

Substitute $$x=2$$ to check the form:

$$\\frac{5(2)^2+40}{2^3-8}=\\frac{20+40}{8-8}=\\frac{60}{0}$$

The numerator approaches a nonzero constant and the denominator approaches $$0$$, so the limit is infinite.

Factor the denominator:

$$x^3-8=(x-2)(x^2+2x+4)$$

The expression becomes:

$$\\frac{5x^2+40}{(x-2)(x^2+2x+4)}$$

Since:

$$5(2)^2+40=60\\neq0$$

and:

$$x^2+2x+4>0$$

near $$x=2$$, the sign of the expression depends on $$x-2$$.

As $$x\\to2^+$$, the expression approaches $$\\infty$$.

As $$x\\to2^-$$, the expression approaches $$-\\infty$$.

Because the one-sided limits are not equal, the two-sided limit does not exist.

Final answer: The limit does not exist.

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$-20$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$\\infty$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$0$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"The limit does not exist.","correcte":true,"explication":"","imageUrl":"$undefined"}]},{"id":"cmpz9r9xo07syjgllmmy6f8m3","texte_question":"

$$\\lim_{x\\to\\infty}\\frac{3-x}{3^x}=?$$

","type_question":"QCM","explication":"

As $$x\\to\\infty$$, analyze the behavior of the numerator and denominator.

The numerator:

$$3-x\\to-\\infty$$

The denominator:

$$3^x\\to\\infty$$

and grows exponentially, much faster than the linear growth of the numerator.

Rewrite the expression:

$$\\frac{3-x}{3^x}=-\\frac{x-3}{3^x}$$

Since exponential growth dominates polynomial growth:

$$\\frac{x-3}{3^x}\\to0$$

as $$x\\to\\infty$$.

Therefore:

$$-\\frac{x-3}{3^x}\\to0$$

Final answer: $$0$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$0$$","correcte":true,"explication":"","imageUrl":"$undefined"},{"texte":"$$$\\frac{1}{3}$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$\\infty$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$-1$$","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmpz9r9xq07t4jgllfybzjpsf","texte_question":"

$$\\lim_{x\\to\\infty}\\frac{5-x}{4^x}=?$$

","type_question":"QCM","explication":"

As $$x\\to\\infty$$, analyze the behavior of the numerator and denominator.

The numerator:

$$5-x\\to-\\infty$$

The denominator:

$$4^x\\to\\infty$$

and grows exponentially, much faster than the linear growth of the numerator.

Rewrite the expression:

$$\\frac{5-x}{4^x}=-\\frac{x-5}{4^x}$$

Since exponential growth dominates polynomial growth:

$$\\frac{x-5}{4^x}\\to0$$

as $$x\\to\\infty$$.

Therefore:

$$-\\frac{x-5}{4^x}\\to0$$

Final answer: $$0$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$\\frac{1}{4}$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$\\infty$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$-1$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$0$$","correcte":true,"explication":"","imageUrl":"$undefined"}]},{"id":"cmpz9r9xt07tajgll39o2f651","texte_question":"

$$\\lim_{x\\to0}\\frac{\\sin(7x)}{x}=?$$

","type_question":"QCM","explication":"

Use the standard limit:

$$\\lim_{u\\to0}\\frac{\\sin(u)}{u}=1$$

Let:

$$u=7x$$

Then:

$$\\sin(7x)=\\sin(u)$$

and:

$$x=\\frac{u}{7}$$

Substitute into the expression:

$$\\frac{\\sin(7x)}{x}=\\frac{\\sin(u)}{\\frac{u}{7}}$$

Simplify:

$$\\frac{\\sin(u)}{\\frac{u}{7}}=7\\cdot\\frac{\\sin(u)}{u}$$

Now take the limit as $$x\\to0$$, which implies $$u\\to0$$:

$$7\\cdot1=7$$

Final answer: $$7$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$\\frac{1}{7}$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$7$$","correcte":true,"explication":"","imageUrl":"$undefined"},{"texte":"$$$14$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$1$$","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmpz9r9xx07tgjgllo9ex0vqx","texte_question":"

$$\\lim_{x\\to0}\\frac{\\sin(4x)}{5x}=?$$

","type_question":"QCM","explication":"

Use the standard limit:

$$\\lim_{u\\to0}\\frac{\\sin(u)}{u}=1$$

Rewrite the expression:

$$\\frac{\\sin(4x)}{5x}=\\frac{4}{5}\\cdot\\frac{\\sin(4x)}{4x}$$

Now take the limit as $$x\\to0$$:

$$\\lim_{x\\to0}\\frac{4}{5}\\cdot\\frac{\\sin(4x)}{4x}=\\frac{4}{5}\\cdot1$$

Simplify:

$$\\frac{4}{5}$$

Final answer: $$\\frac{4}{5}$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$\\frac{5}{4}$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$1$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$\\frac{1}{5}$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$\\frac{4}{5}$$","correcte":true,"explication":"","imageUrl":"$undefined"}]},{"id":"cmpz9r9y007tmjgllukqokrfy","texte_question":"

Which of the following statements is correct about the graph of $$y=\\arctan(2x)$$?

","type_question":"QCM","explication":"

The arctangent function satisfies:

$$\\lim_{x\\to\\infty}\\arctan(x)=\\frac{\\pi}{2}$$

and:

$$\\lim_{x\\to-\\infty}\\arctan(x)=-\\frac{\\pi}{2}$$

Replacing $$x$$ by $$2x$$ does not change the horizontal asymptotes, only the horizontal scaling.

So:

$$\\lim_{x\\to\\infty}\\arctan(2x)=\\frac{\\pi}{2}$$

and:

$$\\lim_{x\\to-\\infty}\\arctan(2x)=-\\frac{\\pi}{2}$$

The arctangent function does not have any vertical asymptotes.

Therefore, the graph has horizontal asymptotes at:

$$y=\\pm\\frac{\\pi}{2}$$

Final answer: horizontal asymptotes at $$y=\\pm\\frac{\\pi}{2}$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"Vertical asymptotes at $$x=0$$ and $$x=\\pi$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"Horizontal asymptotes at $$y=\\pm\\frac{\\pi}{2}$$","correcte":true,"explication":"","imageUrl":"$undefined"},{"texte":"Horizontal asymptotes at $$y=0$$ and $$y=\\pi$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"Vertical asymptotes at $$x=\\pm\\frac{\\pi}{2}$$","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmpz9r9y207tsjglllz2i06yf","texte_question":"

Which of the following statements is correct about the graph of $$y=\\frac{x^2-4}{3x-6}$$?

","type_question":"QCM","explication":"

Factor the numerator and denominator:

$$x^2-4=(x-2)(x+2)$$

$$3x-6=3(x-2)$$

Rewrite the function:

$$y=\\frac{(x-2)(x+2)}{3(x-2)}$$

Cancel the common factor $$x-2$$ for $$x\\neq2$$:

$$y=\\frac{x+2}{3}$$

At $$x=2$$, the original function is undefined, but the simplified function gives:

$$y=\\frac{2+2}{3}=\\frac{4}{3}$$

Since the function simplifies and the discontinuity occurs at a single point, the graph has a removable discontinuity at $$x=2$$.

Final answer: an infinite discontinuity at $$x=2$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"has an infinite discontinuity at $$x=2$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"has a vertical asymptote at $$x=2$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"has a horizontal asymptote at $$y=\\frac{2}{3}$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"has a removable discontinuity at $$x=2$$","correcte":true,"explication":"","imageUrl":"$undefined"}]}]},"categories":["Limits & Continuity"],"date":"2026-06-04T09:04:33.788Z","modified":"2026-06-04T09:04:33.788Z"}}],["$","div",null,{"className":"container mx-auto px-4 pb-12","children":["$","$L18",null,{"targetType":"quiz","targetSlug":"Limits&Continuity-Quiz 3"}]}]]}]}]]

Quiz 3

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