Find
lim
ℎ
→
0
(
sin
⁡
(
3
(
𝑥
+
ℎ
)
2
)
−
sin
⁡
(
3
𝑥
2
)
ℎ
)
.

Question

Find 
lim
ℎ
→
0
(
sin
⁡
(
3
(
𝑥
+
ℎ
)
2
)
−
sin
⁡
(
3
𝑥
2
)
ℎ
)
.

BlackTom AI · Accepted Answer

We are given the limit problem: find lim_{h -> 0} ( sin(3(x+h)^2) − sin(3x^2) ) / h.
Option 1: sin(3x^2). This option suggests the limit would equal sin(3x^2), which is not consistent with the difference quotient form. It ignores the change in the sine argument with respect to h and does not match the derivative structure here.
Option 2: cos(6x). This choice resembles the derivative of sin(6x) with respect to x, but our inner function is 3(x+h)^2, not 6x, and the limit requires a factor of x in the final expression. It misrepresents the chain rule effect and the x-dependence.
Option 3: 6x cos(3x^2). This is the correct form by applying the limit definition of derivative: the numerator is a difference of sin with an inner function varying with h, so the limit equals the derivative of sin(3x^2) with respect to x, which is cos(3x^2) times the derivative of 3x^2, i.e., 6x. Therefore the limit is 6x cos(3x^2).
Option 4: 1. This would indicate a unit limit, which would only occur in very specific normalized cases. Here the rate of change of the inner sine argument is not 1, so this is incorrect.
Option 5: 0. A zero limit would imply no change as h approaches 0, which is not the case since the inner sine argument changes with h and the derivative is generally nonzero when x ≠ 0 (and even when x = 0 the derivative evaluates to 0, but the general expression is not identically 0).
Option 6: 1/3. This fraction does not arise from the derivative of sin(3x^2) or from any standard limit manipulation of this form; there is no reason for the limit to simplify to 1/3, so this is incorrect.

In summary, only Option 3 aligns with the chain-rule differentiation of sin(3x^2) with respect to x, yielding 6x cos(3x^2). The other options misinterpret the role of the inner function, misapply the derivative rules, or ignore the x-dependence intrinsic to the problem.

Find lim ℎ → 0 ( sin ⁡ ( 3 ( 𝑥 + ℎ ) 2 ) − sin ⁡ ( 3 𝑥 2 ) ℎ ) .单项选择题

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