Q.If $\lim_{x \to -1} 4x + k = 6$ then find the value of $k$.

Direct substitution: $\lim_{x\to4}\sqrt{4x+9}=\sqrt{25}=5$.

Since $4x+9>0$ near $x=4$, the square-root function is continuous there and the limit is obtained by direct substitution: …

Substituting $x = -3$ into $y = 10 - 3x$ gives $y = 19$ — option (c).

GSEB Class-12 Statistics, Limit chapter:

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Substituting $x = -1$: $-4 + k = 6$, so $k = 10$.

GSEB Class-12 Statistics, Limit chapter:

Since $4x + k$ is a linear (continuous) function, the limit equals its value at $x = -1$:

$\lim_{x \to -1}(4x + k) = 4(-1) + k = -4 + k$

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Direct substitution gives $\lim_{x\to3}(3x-1)=3(3)-1=8$. Option (D).

Since $3x-1$ is a continuous polynomial function, the limit at $x=3$ is obtained by direct substitution: …

The limit of a constant is that constant, so $\lim_{x\to -2} 10 = \mathbf{10}$.

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Using the standard limit $\lim_{x\to a}\dfrac{x^n-a^n}{x-a}=n\,a^{n-1}$ with $n=4,\ a=3$: value $=4\cdot 27 = \mathbf{108}$.

Given: $\lim_{x\to 3}\dfrac{x^4-81}{x-3}$. At $x=3$ it is the indeterminate form $\dfrac{0}{0}$.

Method (factorisation). Note $81 = 3^4$, so

$x^4-81 = (x^2-9)(x^2+9) = (x-3)(x+3)(x^2+9).$ …

Direct substitution: $\lim_{x\to 3}(3x-1) = 3(3)-1 = 8$ — option (d).

A polynomial function is continuous for all real $x$, so its limit equals its value at the point: …