The limit of a function shows the value of a function when its input approaches a particular value.

Last time, I introduced limits. Today, we will be looking at some complex problems on limits. 

Forthwith, let's get started

Question 1

Evaluate $\lim_{x \to 0}(\frac{\cos 7x-1}{x\sin x})$


If we insert 0 in the denominator, we will get an undefined fraction and this is not factorizable. So we use the l'hopital rule which states that we should take the derivative of the numerator and denominator.

The derivative of the numerator ($\cos 7x-1$) using chain rule is $-7\sin x$

The derivative of the denominator $x\sin x$ using product rule is $\sin x-x\cos x$, Therefore

$$\lim_{x \to 0}(\frac{-7\sin 7x}{\sin x-x\cos x})$$

However, we cannot use direct substitution yet. If we insert 0, we would still get an undefined fraction, hence we apply the l'hopital rule again

The derivative of the numerator $(-7\sin 7x)$ using chain rule is $-49 \cos 7x$ 

The derivate of the denominator is $(\cos x+\cos x-\sin x)$

$$\lim_{x \to 0}(\frac{-49\cos 7x}{\cos x+ \cos x -sinx})$$

$$\lim_{x \to 0}(\frac{-49\cos 7x}{2 \cos x -sinx})$$

Now, we can use direct substitution

$$\lim_{x \to 0}(\frac{-49\cos(7(0))}{2 \cos (0) -\sin(0)})$$

$$\lim_{x \to 0}(\frac{-49}{2})$$

Question 2

Evaluate $\lim_{x \to \infty}(1+\frac{1}{x^{1.5}}+\frac{1}{x^2})$


There is no need to use the l'hopital rule here, we simply use direct substitution

$$\lim_{x \to \infty}(1+\frac{1}{(\infty)^{1.5}}+\frac{1}{(\infty)^2})$$

Any non-zero number raised to power $\infty$ is $\infty$

$$\lim_{x \to \infty}(1+\frac{1}{\infty}+\frac{1}{\infty})$$

Any constant divided by $\infty$ is 0

$$\lim_{x \to \infty}(1+0+0)$$

$$\lim_{x \to \infty}1$$

Question 3

Evaluate $\lim_{x \to 1}(\frac{\sqrt{x+3}-2\sqrt{x}}{\sqrt{x}-1})$


We can rewrite the above expression as

$$\lim_{x \to 1}(\frac{(x+3)^½-2x^½}{x^½-1}$$

If we use direct substitution, we would get an undefined fraction, So, let Apply the l'hopital rule by taking the derivative of the numerator and the denominator

$$\lim_{x \to 1}=\frac{\frac{1}{2}(x+3)^{-½}-2(\frac{1}{2}(x)^{-½})}{\frac{1}{2}(x)^{-½}}$$

$$\lim_{x \to 1}=\frac{\frac{1}{2\sqrt{x+3}}-2(\frac{1}{2\sqrt{x}})}{\frac{1}{2\sqrt{x}}}$$

Now, let use direct substitution

$$\lim_{x \to 1}=\frac{\frac{1}{2\sqrt{1+3}}-2(\frac{1}{2\sqrt{1}})}{\frac{1}{2\sqrt{1}}}$$

$$\lim_{x \to 1}=\frac{\frac{1}{4}-1}{\frac{1}{2}}$$

If we take the L.C.M of the numerator, it will work out to:

$$\lim_{x \to 1}=\frac{\frac{-3}{4}}{\frac{1}{2}}$$

$$\lim_{x \to 1}=\frac{-3}{4}÷\frac{1}{2}$$

$$\lim_{x \to 1}=\frac{-3}{4}\times 2=\frac{-3}{2}$$

Question 4

Evaluate $\lim_{x \to 4 }(\frac{\sqrt{x³}+2\sqrt{x}-12}{x-4})$


The above expression can be rewritten as

$$\lim_{x \to 4}(\frac{x^{\frac{3}{2}}+2x^½-12}{x-4})$$

Applying l'hopital rule

$$\lim_{x \to 4}(\frac{\frac{3}{2}(x^½)+2(\frac{1}{2}x^{-½})}{1})$$

$$\lim_{x \to 4}(\frac{3\sqrt{x}}{2}+2(\frac{1}{2\sqrt{x}}$$

By direct substitution

$$\lim_{x \to 4}(\frac{3\sqrt{4}}{2}+2(\frac{1}{2\sqrt{4}}$$

$$\lim_{x \to 4}(\frac{6}{2}+\frac{2}{4}))=\frac{7}{2}$$

That will be all for now. But before you go...

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