5.10 Limits  (Page 5/5)

The limit has ∞/∞ indeterminate form. Dividing each term of numerator and denominator by n,

$$\Rightarrow \frac{n}{n+1}=\frac{\frac{n}{n}}{\frac{n}{n}+\frac{1}{n}}=\frac{1}{1+\frac{1}{n}}$$

As n→∞, 1/n→0. Hence,

$$\Rightarrow \underset{n\to \infty }{\overset{}{\lim }}\frac{n}{n+1}=1$$

The limit has determinate form. Since cosx is a continuous function, its limit is equal to its value at x=0 i.e. cos0 = 1. Hence,

$$\Rightarrow \underset{x\to 0}{\overset{}{\lim }}\cos x=1$$

The limit has determinate form. Here, ${\log }_{0.5}x$ is a continuous function. The base of exponential function is less than 1. As x approaches zero, the function approaches positive infinity (refer its graph).

$$\Rightarrow \underset{x\to 0}{\overset{}{\lim }}\log {}_{0.5}x=\infty$$

Note that we are not testing limit at singularity. The function is determinate form. Hence, limit is :

$$\Rightarrow \underset{x\to 0}{\overset{}{\lim }}\frac{1}{x-1}=\frac{1}{0-1}=-1$$

The limit has ∞/∞ indeterminate form. Simplifying,

$$\Rightarrow \underset{x\to \infty }{\overset{}{\lim }}1=1$$

The limit has 0/0 indeterminate form. For $x<\mathrm{0,}\left|x^3\right|=-x^3$ . Hence,

$$\Rightarrow \frac{\left|x^3\right|}{x}=-\frac{x^3}{x}=-x^2$$

$$\Rightarrow \underset{x\to 0-}{\overset{}{\lim }}-x^2=0$$

For $x>\mathrm{0,}\left|x^3\right|=x^3$ . Hence,

$$\Rightarrow \frac{\left|x^3\right|}{x}=\frac{x^3}{x}=x^2$$ $$\Rightarrow \underset{x\to 0+}{\overset{}{\lim }}{x}^2=0$$

Clearly, $L_l=L_r=L$ . Thus,

$$\Rightarrow \underset{x\to 0}{\overset{}{\lim }}\frac{\left|x^3\right|}{x}=0$$

The limit has 0/0 indeterminate form. For $x<\mathrm{1,}|x-1|=-\left(x-1\right)$ . Hence,

$$\frac{\left(x^2-1\right)}{|x-1|}=\frac{\left(x^2-1\right)}{-\left(x-1\right)}=-\left(x-1\right)$$ $$\Rightarrow \underset{x\to 1}{\overset{}{\lim }}\frac{\left(x^2-1\right)}{|x-1|}=\underset{x\to 1}{\overset{}{\lim }}-\left(x+1\right)=-2$$

For $x>\mathrm{1,}|x-1|=\left(x-1\right)$ . Hence,

$$\frac{\left(x^2-1\right)}{|x-1|}=\frac{\left(x^2-1\right)}{\left(x-1\right)}=\left(x-1\right)$$ $$\Rightarrow \underset{x\to 1+}{\overset{}{\lim }}\frac{\left(x^2-1\right)}{|x-1|}=\underset{x\to 1+}{\overset{}{\lim }}\left(x+1\right)=2$$

Clearly, $L_l\ne L_r$ . Hence, given limit does not exists.