Q.Compute $\lim_{x \to 0}\dfrac{1 - \cos 2mx}{\sin^2 nx}$ $(m, n \in Z)$.

The limit equals $2$.

The function $\frac{x^2+1}{x}$ is continuous everywhere except $x=0$, so at $x=1$ the limit is simply the value of the function at $x=1$:

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$\lim_{x\to4}(4x^2+2x)=72$.

A polynomial is continuous everywhere, so the limit equals direct substitution: …

$\lim_{x\to1}(1+x+x^2+\ldots+x^{10})=11$.

The expression $1+x+x^2+\ldots+x^{10}$ is a polynomial with $11$ terms ($x^0$ through $x^{10}$), so it is continuous and the limit is found by direct substitutio …

$\lim_{x\to0}(7x^2-2x^3+4x+9)=9$.

By direct substitution (the function is a polynomial, continuous everywhere): …

$\displaystyle\lim_{x\to-1}\dfrac{x^{10}+x^5+1}{x-1} = -\dfrac12$.

The function $\dfrac{x^{10}+x^5+1}{x-1}$ is a rational function, and it is continuous (defined) at $x=-1$ since the denominator $x-1=-1-1=-2\ne0$ there. So the limit is simply the value of the function at $x=-1$ — direct substitution is valid.

Numerator at $x=-1$: $(-1)^{10}+(-1)^5+1 = 1-1+1 = 1$.

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For a polynomial, the limit at a point equals the value of the function there.

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The statement is False; the correct limit value is $12$.

Since $x(x+1)$ is a polynomial, it is continuous everywhere, so the limit as $x\to3$ equals the value of the function at $x=3$:

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