📝 Exponential growth using L'Hopital (15 MCQs)

📖 Explanation: Because the exponential function grows faster than any polynomial, repeated use of L’Hôpital’s rule reduces the numerator’s degree until a constant remains while the denominator stays exponential. The resulting limit is a constant divided by an unbounded exponential, which tends to zero.

📖 Explanation: Differentiating \x^{n}\ reduces the exponent by one each time: the first derivative is \n x^{n-1}\, the second is \n(n-1)x^{n-2}\, and so on. After \n\ differentiations the factor \x\ disappears, leaving the product \n(n-1)\\dots 1=n!\, a constant.

📖 Explanation: The exponential function \e^{x}\ increases proportionally to its current value, causing its growth to accelerate without bound. In contrast, the polynomial \x^{5}\ increases at a rate that depends on a fixed power. Consequently, for sufficiently large \x\, \e^{x}\ exceeds \x^{5}\ and dominates.

📖 Explanation: If the ratio \x^{n}/e^{x}\ tends to zero, its reciprocal \e^{x}/x^{n}\ must diverge to infinity, because both numerator and denominator are positive for large \x\. This follows directly from the definition of limits: a vanishing denominator forces the reciprocal to grow without bound.

📖 Explanation: As \x\ approaches zero from the right, the factor \x\ tends to zero while \\\ln x\ tends to \-\\infty\. The product therefore has the form zero times an infinite magnitude, written as \0\\cdot\\infty\, which is indeterminate because the competing tendencies may produce any finite limit.

📖 Explanation: The first differentiation gives \\\frac{2x}{e^{x}}\. Differentiating again yields \\\frac{2}{e^{x}}\. This expression still contains the exponential denominator, which grows without bound, so the limit of \\\frac{2}{e^{x}}\ as \x\\to\\infty\ is zero.

📖 Explanation: Each application of L’Hôpital’s rule differentiates the denominator, reducing its polynomial degree by one while leaving the numerator unchanged as \e^{x}\. After \n\ applications, the denominator becomes a constant, and the numerator is still \e^{x}\; the constant factor \n!\ appears from differentiating the polynomial, giving \n!\\,e^{x}\.

📖 Explanation: The key property is that the derivative of \e^{x}\ is \e^{x}\ itself, so the function’s growth rate is always a fixed multiple of its current value. This self‑reinforcing behavior means that as \x\ grows, the exponential’s increase outpaces any fixed‑power polynomial, whose derivative eventually becomes negligible.

📖 Explanation: Applying L’Hôpital’s rule repeatedly reduces the polynomial numerator’s degree while the denominator remains exponential. After seven differentiations, the numerator becomes a constant (the 7th derivative of \x^{7}\ is \7!\), whereas the denominator is still \e^{2x}\ multiplied by a power of 2. The constant divided by an unbounded exponential tends to zero.

📖 Explanation: The limit \\\frac{\\ln x}{x}\ is of the indeterminate form \\\frac{\\infty}{\\infty}\. Applying L’Hôpital’s rule once gives \\\frac{1/x}{1}=\\frac{1}{x}\, whose limit as \x\\to\\infty\ is zero. This demonstrates that the logarithm grows much more slowly than any positive power of \x\, consistent with the exponential dominance property.

📖 Explanation: Because the exponent in \e^{x^{2}}\ is \x^{2}\, which itself tends to infinity, the function behaves like an exponential of an ever‑increasing argument. Any polynomial, such as \x^{100}\, can be written as \e^{100\\ln x}\; since \x^{2}\ eventually dominates \\\ln x\, the exponential term outpaces the polynomial.

📖 Explanation: Each application of L’Hôpital’s rule reduces the degree of the polynomial numerator by one while leaving the exponential denominator unchanged. Starting with a fourth‑degree numerator, four successive differentiations are needed until the numerator becomes a constant (the fourth derivative is \4!\). At that point the limit is a constant over an unbounded exponential, which is zero.

📖 Explanation: The limit \\\sqrt{x}\ as \x\\to0^{+}\ approaches zero directly, without any competing infinite behavior. All other listed limits involve a product or quotient where one factor tends to zero while another tends to infinity, creating the classic indeterminate forms such as \0\\cdot\\infty\ or \\\frac{\\infty}{\\infty}\.

📖 Explanation: When one factor tends to zero and the other to infinity, the overall product can converge to zero, a finite non‑zero number, or even diverge, depending on the relative rates at which the factors approach their limits. This sensitivity makes the form \0\\cdot\\infty\ indeterminate; additional analysis (often via L’Hôpital’s rule) is required.

📖 Explanation: Rewrite the expression as \1-\\frac{x^{n}}{e^{x}}\. The term \\\frac{x^{n}}{e^{x}}\ tends to zero because the exponential dominates any polynomial. Consequently, the whole expression approaches \1-0=1\. This limit illustrates that subtracting a slower‑growing term from an exponential does not affect the dominant behavior of the numerator.